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Thermochemistry

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1. IIT–JEE Syllabus

First and second law of thermodynamics; Internal energy; enthalpy, work and heat; heats of reaction, fusion, and vaporisation, Hess’s law; pressure – volume work.

 

  1. Thermodynamics 

The study of the transformation of energy is called thermodynamics. Actually, it deals with energy in its various forms, which include thermal, chemical, electrical, and mechanical, with the restrictions on the transformation of one type of energy in to the other types and with the relation of energy changes to physical and chemical changes.

2.1 Terminology of Thermodynamics

System and Surroundings

System is part of the universe which is arbitrarily set off from the rest of the universe by definite boundaries for the purpose of experimental or theoretical studies. The remainder of the universe is then, in fact the surroundings of the system. The space separating the system from its surrounding is termed as boundary.

Types of Systems

  1. a) Real system In experimental work, the system is called Real.
  2. b) Ideal system In pencil and paper work, the system treated is called ideal. An ideal system is always considered to simplify the thermodynamic problems.
  3. c) Isolated system A system is said to be isolated when it can neither exchange energy nor matter with its surroundings.
  4. d) Closed system A system is said to be closed when it permits passage of energy but not mass, across the boundary.
  5. e) Open system A system which can exchange both energy and matter with its surroundings.

State of System (State variables)

The quantities whose value serve to describe the system completely are called the thermodynamic properties of the system. Once the properties of the system are completely specified, one says that the state of the system is specified. Thus, the defining properties are sometimes called state variables or state properties. Examples of state properties are pressure, volume, temperature and composition. The question now arises as how many variables must be determined to define the system completely. The answer to this question can be obtained by considering the following example.

A homogeneous system consists of a single substance and hence the composition is fixed automatically. The state of a homogeneous system can, therefore, be defined by only three variables:

  1. i) Pressure
  2. ii) Volume and 

iii) Temperature

For a homogenous system of definite mass, these three properties are related to one another by a mathematical equation PV = RT, called equation of state. With the help of this equation (PV = RT) the values of any one of these properties can be determined knowing the values of the other two properties. Therefore, state of a simple homogeneous system may be completely defined by specifying only two of the three variables (i.e.) pressure, temperature and volume. The two variables generally specified are temperature and pressure. These are termed as independent variables. The third variable, generally volume is called a dependent variable because its value depends upon the values of pressure and temperature.

When we are considering a closed system consisting of one or more components, mass is not a state variable.

In order to define the state of a homogeneous system having more than one substance, one must consider and describe each of the phases of the system. For each phase, one must specify the content (i.e.) the amount of each substance present, and two other independent variables.

In order to define a system completely, the state variables are generally (T), Pressure (P) Volume (V) and concentration (n). Besides these there are two more variables. Work (W) and heat (q), which are not state properties. These six variables play an important role in defining chemical systems completely.

Properties of a System: The observable properties of system are of two types:

  1. a) Extensive Properties: There are some properties called Extensive properties whose values are proportional to the mass of the portion of the system or one can say that extensive properties are dependent upon size of the system.
  2. b) Intensive Properties: There are some properties of a system called intensive properties, whose values are independent of the quantity of matter contained in the system.
Extensive Property Intensive Property
Volume Molar volume
No. of moles Density
Mass Refractive index
Free Energy Surface tension
Entropy Viscosity
Enthalpy Free energy per mole, specific heat
Heat capacity Pressure, Temperature, Boiling Point, Freezing Point

2.2 Internal Energy 

It is the energy associated with a system by virtue of its molecular constitution and the motion of its molecules. The contribution of energy due to molecular constitution is known  as cinternal potential energy and the contribution of energy due to the motion of molecules is called internal kinetic energy. Internal energy of a system is given by the sum of two types of energies.

Determination of ΔE: When a reaction is carried out in such a manner that the temperature and volume of the reacting system remain constant, then the internal energy change (ΔE) of the reaction is equal to the heat exchanged with the surrounding.

2.3 Enthalpy 

When we deal certain process in open vessels (at constant pressure). It becomes essential to introduce in place of internal energy, a new themrodynamic function called heat enthalpy. This new function is denoted by H.

 H = E + PV 

The change in enthalpy of a given system is given as follows,

ΔH = H2 – H1

or ΔH = (E2 + P2V2) – (E1 + P1V1) = (E2 – E1) + (P2V2 – P1V1)

or ΔH = ΔE + ΔPΔV

If P is maintained constant

ΔH = ΔE + PΔV

or ΔH = Q

Hence the change in enthalpy of the system ΔH may be defined as the amount of heat absorbed at constant pressure.

2.4 Thermodynamic Process

  1. i) Adiabatic Process: When a process is carried out under such conditions that no exchange of heat takes place between the system and its surrounding, the process is called adiabatic.

Ex: The sudden bursting of a cycle tube. If the process is exothermic the heat evolved will remain in the system and, therefore, the temperature of the system rises and vice versa.

  1. ii) Isothermal Process: The process which occurs at constant temperature is called isothermal. In an isothermal change the temperature is kept constant by adding heat or taking it away from the substance. So in an isothermal process the system exchanges heat with the surroundings.

iii) Isobaric Process: If the pressure of the system remains constant during each step of the change in the state of a system, this process is said to be an isobaric process.

  1. iv) Isochoric process: A process is defined to be isochoric if the volume of the system remains constant during the process.

2.5 First law of Thermodynamics 

Energy may be converted from one form to another, but it is impossible to create or destroy it. There are various ways of expressing the first law of thermodynamics. Some of the selected statements are given below:

  1. i) When work is transformed into heat or heat into work, the quantity of work is mechanically equivalent to the quantity of heat.
  2. ii) Energy of an isolated system must remain constant, although it may be transformed from one form to another.

iii) Energy in one form, if it disappears will make its appearance in an exactly equivalent in another form.

  1. iv) It is never possible to construct a perceptual motion machine that could produce work without consuming any energy.

2.5.1 Mathematical Formulation of the First Law

Suppose a system absorbs a quantity of heat q and its state change form A to B (fig – I). This heat is used up,

  1. i) In increasing the internal energy of the system i.e, ΔE = EB – EA
  2. ii) To do some external work ‘w’ by the system on its surroundings. 

From the first law, we get.

Heat observed  by the system  = its internal energy + work done by the system.

q = ΔE + w …1

2.6 Differential form of the First Law 

For an infinitesimal process, equation (1) takes the form

δq = dE + δw …2

  1. a) Change in internal energy is independent of the path taken. So in mathematical terms, an exact differential is always denoted by a notation d, e.g. dE in equation (2).
  2. b) q and w are not state functions because changes in their magnitude is dependent on the path by which the change is accomplished. Mathematically q & w are not exact differentials and we always write the inexact – differential by δq, δw etc.
  3. c) For a cyclic process, the change in the internal energy of the system is zero because the system is brought back to the original condition.

= 0

or = i.e. the total work obtained is equal to the net heat supplied.

  1. d) In an isolated system, there is no heat exchange with the surrounding i.e. δq = 0

dE + δW = 0

or δw = – dE

The sign convention: According to latest S.I. convention, w is taken as negative if work is done by the system whereas it is taken as positive if work is done on the system. When heat is given by the system to surrounding it is given as negative sign. When heat is absorbed by the system from the surrounding then positive sign is given.

2.7 Heat, Energy, Work

Heat 

Like work, heat is regarded in thermodynamics as energy in transit across the boundary separating a system from its surroundings. However, quite unlike work, heat transfer results from a temperature difference between system and surroundings, and simple contact is the only requirement for heat to be transferred by conduction. Heat transfer ceases when thermal equilibrium is attained. All other forms of energy can be quantitatively converted into work but heat cannot be completely converted into work without producing permanent changes either in the system or in the surroundings. Heat is an algebraic quantity and is symbolised as Q.

The units of heat are those of work and energy. The sign convention used for a quantity of heat Q is opposite to that used for work. Heat added to a system is given by a positive sign, whereas heat extracted from a  system is given negative sign.

Like work Q is not a state function, its work depends on the path which is followed for carrying out the transformation in the state of the system. Heat is, therefore a path dependent quantity.

Energy

It is the capacity for doing work. Whenever, there is a rearrangement of atoms as in chemical reactions or as in changes in the state of matter (e.g. fusion, vapourisation, sublimation, etc.) energy changes are involved. Energy manifests itself in various forms. For example, the kinetic energy of a body is due to its motion and potential energy of a body is due to its motion and potential energy of a body is due to its position in space. These two forms of energy are little importance for chemical system. The energy which is of great significance from chemical point of view is the internal energy of substance.

The energies acquired by a system in a force field like electric, magnetic, gravitational surface etc. are termed as external energies and are usually not considered as part of the internal energy of the system.

Energy is an extensive property i.e., its magnitude depends upon the quantity of material in the system. 

In SI system, the unit of energy is expressed in J or kJ.

Work

In thermodynamics, work is generally defined as the force multiplied by the distance. If the displacement of body under the force F is ds, the work done will be,

δW = Fds

The symbol δW stands for the small amount of work and also the inexactness of the function. Several things should be noted in the definition of work.

  1. a) Work appears only at the boundary of the system
  2. b) Work appears only during a change in state
  3. c) Work is manifested by an effect in the surroundings
  4. d) Work is an algebraic quantity. It is positive if the work has been produced in the surroundings. It is negative if the work has been destroyed in the surroundings.
  5. e) In SI system of units, work is expressed in Joule or Kilojoule 1J = 1 Nm.

Types of work

  1. a) Gravitational Work: The work is said to be done when a body is raised through a certain height against the gravitational field. Suppose a body of mass m is raised through a height h against the gravitational field. Then the magnitude of the gravitational work is mgh.
  2. b) Electrical work: This type of work is said to be done when a charged body moves from one potential region into another. If the charge is expressed in coulombs and the potential difference in volts, then the electrical work is given by QV.
  3. c) Mechanical Work: work associated with change in volume of a system against an external pressure is referred to as the mechanical or pressure-volume work.

Now we shall discuss mechanical work in detail.

Work has been done by the system if a weight has been raised in the surrounding, work has been done on the system if a weight has been lowered.

2.8 Work in reversible process

  1. a) Expansion of a gas 
  1. Suppose n moles of a perfect gas is enclosed in a cylinder by a frictionless piston. The whole cylinder is kept in large constant temperature bath at T°K. Any change that would occur to the system would be isothermal.Suppose area of cross section of cylinder = a sq. cm

Pressure of the piston  = P

Distance through which gas expands = dll cm

Then force (F) = P ×

Work done by the gas  = F × dl P × a × dl

a × dl  = dV

w = P . dV

Let the gas expand from initial volume V1 to the final volume V2, then the total work done(w)  = 

2.9 Work done in isothermal reversible expansion of an ideal gas 

The small amount of work done, δw when the gas expands through, a small volume dV, against the external pressure, P is given by 

δw = – PdV

Total work done when the gas expands from initial volume V1 to final volume V2 will be W =

Ideal gas equation PV = nRT

i.e, P =

Hence W = [ T = constant] 

W = – nRT ln

= – 2.303 nRT log

= –2.303 nRT log 

The –ve sign indicates work of expansion.

Note: 1 Work in the reversible process is the maximum and is greater than that in the irreversible process.

Illustration 1: 70 gms of nitrogen gas was initially at 50 atm and 25°C

  1. a) It was allowed to expand isothermally against a constant external pressure of one atmosphere. Calculate, ΔU, ΔQ and ΔW assuming the gas to behave ideally.
  2. b) Also find out the maximum work that would be obtained if the gas expanded reversibly and isothermally to one atmosphere. 

Solution: a) Amount of gas n = = 2.5 moles 

Initial pressure; P1 = 50 atm

Final pressure  = 1 atm

Since the gas is ideal, ΔU = 0, as the temperature is constant.

The work obtained Δw = P2 (V2 – V1)

P2  =

=  

= 1.5 Kcal

  1. b) The maximum work for its isothermal reversible expansion.

ΔW = nRT ln

= 2.303 × 2.5 × 2 × 298 log = 5.8 Kcal

Exercise 1: Calculate the maximum work done by the system when 2 mole of ideal gas expands from 2.4 dm3 to 5.6 dm3 at 10°C reversibly. Also calculate the heat change. Repeat the above calculations at 110°C.

2.10 Adiabatic Process (Reversible)

An adiabatic change by definition, is one which does not allow any transfer of heat, i.e., q = 0, it follows from the 1st law,

ΔU = – W

dU  = – dW

Let only mechanical work of expansion or contraction is involved, dW = PdV. Moreover,
dU= CVdT

CVdT = –PdV

For a system of 1 mole of an ideal gas, expanding adiabatically and reversibly from temp T1 to T2 and volume V1 to V2, we have 

CVdT =

or .

or

ln =

or

or = K 

Cp – Cv = R

ln

or 

or (γ-1) ( γ = Cp/Cv)

or

or

Substituting  T by

= constant 

PVγ = R × constant = constant

Similarly = constant

TγP1-γ = constant 

Adiabatic work:

W = – CvT = –Cv(T2–T1) = Cv (T1 – T2)

Where T1,T2 are initial and final temperatures.

For 1 mole of gas T = PV/R

Hence adiabatic work

W = Cv =(P1V1 – P1V2)

or w =

Slope of PV curve in adiabatic & isothermal expansion.

For  isothermal expansion of the gas, PV  = K

The slope of the PV curve will be obtained from

For the adiabatic expansion of the gas PVγ = K

P = K/Vγ ; =

In the both the changes the slope is negative, since γ, is greater than 1, the slope in the adiabatic P – V curve will be large than that in the isothermal one.

Illustration  2: A sample of a gas initially at 27°C is compressed from 40 litres to 4 litres adiabatically and reversibly. Calculate the final temperature (Cv = 5 cal/mole)

Solution: Assuming ideal behaviour 

Cp = Cv + 2 = 7 cal / mole

γ = Cp/Cv  = 7/5

If T be the final temperature thus 

T = 300 × (10)7/5-1 

= 753.6°K

2.11 Heat Capacity 

Need for the function: Heat content or enthalpy function is particularly used to measure heat changes accompanying a process at constant volume. It therefore becomes necessary to introduce a new function called heat capacity which relates the heat changes to the temperature changes at constant pressure or at constant volume.

The heat capacity of a system is defined as the quantity of heat required for increasing the temperature of one mole of a system through 1°C. Heat capacity may be given as follows.

C = …(1)

  1. i) Heat capacity at constant volume

By first law of thermodynamics, 

We have

δq = dE + PdV …(2)

on substituting the value of δq in equation (2) We get

C = …(3)

If the volume is kept constant than

CV = …(4)

Hence the heat capacity at constant volume of a given system may be defined as the rate of change of internal energy with temperature.

  1. ii) Heat capacity at constant pressure

If the pressure is held constant, equation (3), becomes as follows.

Cp =

Or Cp =

Hence the heat capacity at constant pressure of a system may be defined as the rate of change of enthalpy with temperature.

2.12 Limitations of first Law of thermodyanamics

  1. This law fails to tell us under what conditions and to what extent it is possible to bring about conversion of one form of energy into the other.
  2. The first law fails to contradict the existence of a 100% efficient heat engine or a refrigerator.

2.13 Second law of thermodynamics

It has been stated is several forms as follows.

  1. i) All the spontaneous process are irreversible in nature.
  2. ii) It is impossible to obtain work by cooling a body below it lowest temperature

iii) It is impossible to take heat from a hot reservoir and convert it completely into work by a cyclic process without transferring a part of it to a cold reservoirs.

  1. iv) Heat cannot of itself pass from a colder body to hotter body without the intervention of external work.
  2. v) It is impossible to construct a machine functioning in cycle which can convert heat completely into equivalent amount of work without producing change elsewhere.
  3. vi) The entropy of universe is always increasing in the course of every spontaneous process.

vii) Spontaneous or natural process are always accompanied with an increase in entropy.

2.14 Efficiency of a Heat Engine

The relationship between W, the net work done by the system and q2, the quantity of heat absorbed at the higher temperature T2, in case of the cyclic process (i.e. the carnot cycle), can be obtained from the following two equations.

W = R (T2 – T1) ; q2 = RT2; W = q2

The fraction of the heat absorbed by an engine which it can convert into work gives the efficiency(η).

η =

The net heat absorbed by the system, q is equal to q2 – q1 and according to the first law of thermodynamics, this must be equivalent to the net work done by the system. Thus,

W = q2 – q1

or = = η.  Since is invariably less than l, the efficiency of a heat engine is always less than 1.

  1. Thermochemistry

Thermochemistry deals with the transfer of heat between a chemical system and its surroundings when a change of phase or a chemical reaction takes place within the system. In general, a chemical reaction can be either exothermic or endothermic. In the former case, heat is released to the surroundings when the reactants at a given temperature and pressure are converted to the products at the same temperature and pressure, and in the latter heat is absorbed by the system, from the surroundings. 

Sign conventions: If the heat is absorbed by the system (q>0) then the reaction is said to be endothermic  and ΔE or ΔH value is given a positive sign. If the heat is evolved (q<0) the reaction is said to be  exothermic, and ΔE or ΔH values is given negative sign.

Standard States:  In the computation of heat of reactions it is a convention to assume that the heat of formation of elements in their standard states is zero. The standard state is taken as 1 atm pressure and at a constant temperature. Standard states for various forms of matter are summarized below:

State Standard State
Gas  Ideal gas at 1 atm and the given temperature 
Liquid  Pure liquid at 1 atm and the given temperature 
Solid  Stable crsytalline form at 1 atm and given  T (e.g. graphite form of carbon, rhombic form of sulphur)

At standard state the heat of reactions are denoted by or at given temperature.

  1. Various Types of Enthalpies of Reactions 
  2. i) Enthalpy of formation : Enthalpy change when one mole of a given compound is formed from its elements.

H2(g) + 1/2O2(g) ⎯→  2H2O(l), ΔH = –890.36 kJ / mol

  1. ii) Enthalpy of combustion: Enthalpy change when one mole of a substance is burnt in oxygen

CH4 + 2O2(g) ⎯→ CO2  + 2H2O(l), ΔH  = –890.36 kJ / mol

iii) Enthalpy of Neutralization:  Enthalpy change when one equivalent of an acid is neutralized by a base in dilute solution. This is constant and its values is –13.7 kcal for neutralization of any strong acid by a base since in dilute solutions they completely dissociate into ions.

H+ (aq) + OH (aq) ⎯→ H2O(l) ΔH = –13.7 kcal

For weak acids and bases, heat of neutralization is different because they are not dissociated completely and during dissociation some heat is absorbed. So total heat evolved during neutralization will be less.

e.g. HCN + NaOH ⎯→ NaCN + H2O ΔH = –2.9 kcal

Heat of ionization in this reaction is equal to (–2.9 + 13.7) kcal = 10.8 kcal

  1. iv) Enthalpy of hydration: Enthalpy of hydration of a given anhydrous or partially hydrated salt is the enthalpy change when it combines with the requisite no.of mole of water to form a specific hydrate. For example, the hydration of anhydrous copper sulphate is represented by 

CuSO4(s) + 5H2O (l) ⎯→ CuSO45H2O, ΔH° = –18.69 kcal

  1. vi) Enthalpy of Transition: Enthalpy change when one mole of a substance is transformed from one allotropic form to another allotropic form.

C (graphite) ⎯→  C(diamond), ΔH° = 1.9 kJ/mol

  1. Laws of Thermochemistry 

For some reactions it is not convenient to measure the heat change in the laboratory. So conventional procedure based on the principle of conservation of energy has been suggested which can be stated as follows:

  1. The heat of formation of any compound is equal in magnitude and of opposite sign to the heat of dissociation of that compound at the given temperature and pressure.

For example, enthalpy of formation of liquid water from its elements hydrogen and oxygen is –285.830kJ mol–1 and the enthalpy of dissociation is  285.830 kJ mol–1. Thus the process can be represented by 

H2(g) + 1/2  O2(g) ⎯→ H2O (l),             ΔH(298 K) = –285.830 kJ

H2O(l) ⎯→ H2(g) +½ O2(g), ΔH (298 K) = + 285.830 kJ

  1. The total enthalpy change of a reaction is the same, regardless of whether the reaction is completed in one step or in several steps. (Hess’s Law of constant heat summation.) It has been experimentally verified and is also a consequence of the law of conservation of energy. It is of particular utility in calculation of the heats of reactions which are difficult for practical calorimetric measurements.

For example : Carbon be converted into CO2 in 1 step 

C(s) + O2(g) ⎯→ CO2(g) ΔH = –94 kcal

Or in two steps 

C(s) + 1/2  O2(g) ⎯→ CO(g) ΔH1 = –26.,4 kcal

CO)(g) + 1/2  O2(g) ⎯→ CO2(g) ΔH2 = –67.6 kcal

According to Hess’s law: ΔH = ΔH1 + ΔH2  = –26.4 –67.6=94 Kcal

  1. Lattice Energy of an Ionic Crystal (Born–Haber Cycle)

The change in enthalpy that occurs when 1 mole of a solid crystalline substance is formed from its gaseous ions.

Step  1: Conversion of metal to gaseous atoms 

M(s) ⎯→ M(g) , ΔH1 = sublimation 

Step 2: Dissociation of X2 molecules to X atoms

X2(g) ⎯→ 2X (g), ΔH2 = Dissociation energy 

Step 3: Conversion of gaseous metal atom to metal ions by losing electron

M(g) ⎯→ M+ (g) +  e , ΔH3 =  Ionization energy 

Step 4: X(g) atoms gain an electron to form M ions

X(g) + e ⎯→ X(g) , ΔH4 = Electron affinity

Step 5: M+ (g) and X (g) get together and form the crystal lattice

M+ (g) + X (g) ⎯→ MX(s)   ΔH5 = lattice  energy 

Applying Hess’s law we get 

ΔH1 + 1/2 ΔH2 + ΔH3 + ΔH4 + ΔH5 = ΔHf (MX)

On putting the various known values, we can calculate the  lattice energy.

  1. Bond Energies 

Whenever a chemical bond is formed energy is released. Conversely, energy is required to rupture a chemical bond. The energy required to break a particular bond in a gaseous molecule is referred to as bond-dissociation energy.

H–OH(g) ⎯→2H(g) + ½O(g) ΔH = 498 kJ

O–H(g) ⎯→H2(g) + ½O2 (g) ΔH = 430 kJ

In such case, bond energy is expressed as the average of the bond dissociation energies of various similar bonds   ΔHO–H = (498 + 430)/2 = 464 kJ mol–1

Hence, bond energy may be defined as the average amount of energy required to break one mole bonds of that type in gaseous molecules.

Thus, the values given in bond energy data can help us in: 

  1. i) Calculating standard enthalpy of reactions
  1. Calculation of bond energies of some specific bond in the molecule 
  1. Variation Of Heat Of Reaction With Temperature 

The heat of reaction depends on the temperature. The relation between the two is known as Kirchoff’s equation.

  1. i) = ΔCP
  2. ii) = ΔCV

ΔCP = molar heat capacity of products – molar heat capacity of reactants (at constant pressure)

ΔCV = molar heat capacity of products – molar heat capacity of reactants (at constant volume)

  1. Bomb Calorimeter 

The calorimeter used for determining enthalpies of combustion known as the bomb calorimeter is shown as Figure. 

This apparatus was devised by Berthelot (1881) to measure the heat of combustion of organic compounds. A modified form of the apparatus shown in Figure consists of a sealed combustion chamber, called a bomb, containing a  weighed quantity of the substance in a dish along with oxygen under about 20 atm pressure. The bomb is lowered in water contained in an insulated copper vessel. This vessel is provided with  a stirrer and a thermometer reading up to 1/100th of a degree. It is also surrounded by an outer jacket to ensured complete insulation from the atmosphere. The temperature of water is noted before the substances  is ignited by an electric current. After combustion, the rise in temperature of the system is noted on the thermometer and heat of combustion can be calculated from the heat gained by water and the calorimeter.

By knowing the heat capacity of calorimeter and also the rise in temperature, the heat of combustion can be calculated by using the expression 

Heat exchange = Z × ΔT

Z–Heat capacity of calorimeter system 

ΔT– rise in temp.

Heat changes at constant volumes are expressed in ΔE and Heat changes at constant pressure are expressed in ΔH.

Also, ΔH = ΔE + ΔnRT

Δn =  gaseous product moles – gaseous reactant moles.

Exercise 2: When 1-pentyne (A) is treated with 4N alcoholic KOH at 175°C, it is converted slowly into an equilibrium mixture of 1.3% 1-pentyne (A), 95.2% 2-pentyne (B) and 3.5% of 1,2-pentadiene (C). The equilibrium was maintained at 175°C. Calculate ΔG0 for the following equilibria.

B ==A ΔG10 = ?

B ==C ΔG20 = ?

From the calculated value of ΔG10 and ΔG20 indicate the order of stability of A, B and C. Write a reaction mechanism showing all intermediates leading to A, B and C [10]

 

Exercise 3: Two moles of a perfect gas undergo the following processes:

  1. a) a reversible isobaric expansion from (1.0 atm, 20.0L) to
    (1.0 atm, 40.0 L);
  2. b) a reversible isochoric change of state from (1.0 atm, 40.0 L) to (0.5 atm, 40.0 L);
  3. c) a reversible isothermal compression from (0.5 atm, 40.0 L) to
    (1.0 atm, 20.0 L).
  4. i) Sketch with labels each of the processes on the same P-V diagram.
  5. ii) Calculate the total work (w) and the total heat change (q) involved in the above processes.

iii) What will be the values of ΔU, ΔH and ΔS for the overall process?

  1. Solution to Exercises

Exercise 1: W = nRT ln; W = (2) (8.314) (283) ln = 3987 J

According to first law of Thermodynamics; Q = ΔE + W

Since ΔE = 0 at constant temperature; Q = W; Q = 3987 J

When T2 = 273 + 100 = 383 K

W = (2) (8.314) (383) ln = 5396 J; Similarly Q = 5396 J

Exercise 2: B = A, ΔG10 = {( 2.303× 8.314×448)log(1.3/95.2)}=15.998 kJ

B = C, ΔG20 = {( 2.303× 8.314×448)log(3.5/95.2)}=12.305 kJ . 

Stability order B > C > A

Exercise 3:
ii) Total work (W) = W1 + W2 + W3

= – PΔV + 0 + 2.303 nRT log

= –1 × 20 + 2.303 × 2 × 0.082 × 121.95 log2

= – 20 +13.86 = –6.13 L atm

Since the system has returned to its initial state i.e. the process is cyclic so ΔU = 0

ΔU = q + W = 0, so; q = – W = 6.13 L. atm = 620.7 J

In a cyclic process heat absorbed is completely converted into work

iii) Entropy is a state function and since the system has returned to its initial state so ΔS = 0. Similarly ΔH = 0 and ΔU = 0 for the same reason i.e. U and H are also state functions having definite values in a given state of a system.

  1. Solved  Problems 

11.1 Subjective 

Problem 1: A Carnot’s engine works between 120°C and 30°C. Calculate the efficiency. If the power produced by the engine is 400 watts, calculate the heat absorbed from the source and rejected to the sink every second.

Solution: Efficiency, η =

Here T1 = 273 + 120 = 393 K

T2 = 273 + 30 = 303 K

η = = 0.229 = 22.9%

Again η =

Thus, heat absorbed from the source,

Q1 = = 1747 watts.

Also

The heat rejected to the sink

Q2 = Q1 ×

= 1747 × = 1347 watts

Problem 2: The enthalpy of neutralisation of ammonium hydroxide by hydrochloric acid is 51.46 kJ mol–1. Calculate the enthalpy of ionisation of ammonium hydroxide.

Solution: i) NH4OH (aq) + H+ (aq) ⎯→ (aq) +_ H2O (l); ΔH° = –51.46 kJ mol–1

  1. ii) H+(aq) + OH (aq) ⎯→ H2O(l); ΔH° = – 57.32 kJ mol–1

iiii) NH4OH (aq) ⎯→ (aq) + OH (aq); ΔH0 = ΔH0 ionisation

On adding (ii) and (iii), we get

NH4OH (aq) + H+ (aq) ⎯→ (aq) + H2O(l)

or Δ H0 = –57.32 + Δionisation

From equation (i), we have

ΔH0 = –51.46 kJ mol–1

Hence (–57.32 + ΔH0ionisation) kJ mol–1 = –51.46 kJ mol–1

or ΔH0Ionisation = + 5.86 kJ mol–1.

Problem 3: If EC – C is 344 kJ mole–1 and EC – H is 415 kJ mole–1, calculate the heat of formation of propane. The heats of atomization of carbon and hydrogen are 716 kJ mole–1 and 433 kJ mole–1 respectively.

Solution: The heat of formation is the sum of the heats of atomization and bond energies. For propane, the heats of atomization are

3C(s) = 3C(g0; ΔH = 3 × 716 = 2148 kJ

4H2,(g) = 8H(g); ΔH = 4× 433 = 1732 kJ

The bond energies are

2EC – C = 2 × –344 = –688 kJ

8EC – H = 8 × –415 = –3320 kJ

Adding, 

3C + 4H2 = C3H8; ΔHf = 2148 + 1732 – 688 – 3320 = –128 kJ mole–1

Problem 4: Calculate the amount of heat evolved during the complete combustion of 100 ml of liquid benzene from the following data:

  1. i) 18 gm graphite on complete combustion evolves 590 KJ of heat

iii) 15889 kJ of heat is required to dissociate all the molecules of 1 litre water into H2 & O2.

iii) The heat of formation of liquid benzene is +50 kJ/mole.

  1. iv) Density of C6H6(l) = 0.87 gm ml–1

Solution: C(s) + O2(g) ⎯→ CO2(g)

(CO2) = kJ /mol = 393.33 kJ/mol

H2(g) + ½ O2(g) ⎯→ H2O (l)

(H2O (l)) = – = –286 kJ mol–1

C6H6(l) + 7.5 O2(g) ⎯→ 6CO2(g) + 3H2O(l)

= 6 × (–393.33) + 3(-286) – 50 = –2359.98 – 858 – 50 –3268 kJ/mol.

Heat evolved = 3268 × kJ = 3645 kJ.

Problem 5: Calculate the enthalpy of vaporization for water from the following 

H2(g) + 1/2 O2(g) ⎯→ H2O (g) ΔH = – 57.0 kcal

H2(g) + 1/2 O2(g) ⎯→ H2O(l) ΔH = – 68.3 kcal

Also calculate the heat required to change 1 gm H2O (l) to H2O (g) 

Solution: H2(g) + 1/2 O2(g) ⎯→ H2O (g) ΔH = –57.0 kcal —————– (1)

H2(g) + 1/2 O2(g) ⎯→ H2O (l) ΔH = – 68.3 kcal——————(2)

Subtracting (2) from (1) 

H2O (l) ⎯→ H2O (g) ; ΔH = 11.3 kcal

Enthalpy of vaporization for H2O = 11.3 kcal

Also 18 g H2O requires enthalpy of vaporization  = 11. 3 kcal

1 g H2O requires kcal = 0.628 kcal

Problem 6: The standard enthalpy of combustion of H2, C6H10 and Cyclohexane (C6H12) are – 241, – 3800, – 3920 kJ mole–1 at 25°C respectively. Calculate the heat of hydrogeneation of cyclohexane.

Solution: We have to find Δ H for 

C6H10 + H2 ⎯→ C6H12

Given H2 + 1/2 O2 ⎯→ H2O ΔH = – 241 kJ …(1)

C6H10 + O2 ⎯→ 6CO2 + 5H2O ΔH = – 3800 kJ …(2)

C6H12 + 9O2 ⎯→6CO2 + 6H2O ΔH = – 3920 kJ …(3)

Adding equation (1) and (2) and then subtracting equation  3 

C6H10 + H2 ⎯→ C6H12 ΔH = – 121 kJ

Heat of hydrogenation of cylohexene  = – 121 kJ

Problem 7: Calculate the value of for OH(aq.) if

(H2O(l)) = –68 kcal/mol

(H+(aq.)) = 0

Solution: Given

  1. i) H2(g) + O2(g) ⎯→ H2O(l) ΔH1 = –68.7 kcal/mol
  2. ii) ⎯→ H+(aq.) + e ΔH2 = 0

Known

iii) H+(aq.) + OH(aq.) ⎯→ H2O(l) ΔH3 = –13.7 kcal/mol

Consider

  1. iv) ⎯→ OH(aq.)
  2. iv) will be obtained by applying (i) – (ii) – (iii)

= –68.7 + 13.7 = –55 kcal mol–1

Problem 8: The internal energy change in the conversion of 1.0 mole of the calcite form of CaCO3 to the aragonite form is + 0.21 kJ. Calculate the enthalpy change when the pressure is 1.0 bar. Given that the densities of the solids are 2.71 g cm–3 and 2.93 g cm–3 respectively.

Solution: ΔH = ΔE + PΔV

Given ΔE = +0.21 kJ mole–1 = 0.21 × 103 mole–1

P = 1 bar = 1 × 105 pa

ΔV = V(aragonite)V(Calcite)

= cm3 mole–1 of CaCO3

= – 2.77 cm3 = – 2.77 × 10–6 m3

ΔH = 0.21 × 103 – 1 × 105 × 2.77 × 10–6

= 209.72 J

= 0.20972 kJ mole–1

Problem  9: 3.67 litres of ethylene and methane gaseous mixture on complete combustion at 25°C produces 6.11 litre of CO2. Find out the amount of heat evolved on burning one litre of gaseous mixture. The heats of combustion of C2H4 and CH4 are – 1423 and – 891 kJ mole–1 at 25°C.

Solution: Let volume of C2H4 in the mixture = x litre 

Volume of CH4 in the mixture  = ( 3.67 – x) Lit 

C2H4 + 3O2 ⎯→ 2CO2 + 2H2O

CH4  + 2O2 ⎯→ CO2  + 2H2O

From above two equations, it is clear that x litre C2H4 will from 2x litre CO2 and (3.67–x) litre CH4 will from ( 3.67 –x) litre CO2

Hence, 2x + 3.67 – x = 6.11 

x + 3.67 = 6.11

x = 2.44

Volume of CH4 percent in the mixture  = 3.67 – 2.44  = 1.23 litre 

Now, volume of C2H4 in one litre mixture =  =  0.6648 litre 

Volume of CH4  in one litre mixture =

= 0.3352 litre 

Above volumes are there at 25°C

Number of moles of methane  PV = nRT

1 × 0.3352 = n × 0.0821 × 298

= n = 0.0137 mole 

Heat evolved by combustion of methane 

= Number of methane moles × Heat of combustion 

= 0.0137 × 891 = 12.21 kJ

Number of moles of ethylene 

PV = nRT

1 × 0.6648 = n × 0.0821 × 298

n = 0.02717 mole 

Heat evolved by combustion of ethylene 

= 0.02717 × 1423 = 38.68 kJ

Total heat evolved  = 38.68 + 12.21

= 50.89 = 50.9 kJ.

Problem 10: Calculate the proton affinity of NH3(g) from the following data (in kJ / mole)

Δdissociation : H2(g) = 218

Δformation : NH3(g) = – 46

Lattice energy of NH4Cl = – 683

Ionisation energy of H = 1310

Electron affinity of Cl = 348

Bond dissociation energy Cl2(g) = 124

Lattice energy NH4Cl(s) = – 314

Solution: We have to calculate ΔH for the following equation 

NH3(g) + H+(g) ⎯→ (g) ————- (1) 

Given; H2(g) ⎯→ 2H(g) : ΔH1 = 218 kJ / mole 

N2(g) + H2(g) ⎯→ NH3(g) : ΔH2 = – 46 kJ / mol 

NH4Cl (s) ⎯→ (g) + Cl(g) : ΔH3 = + 683 kJ / mol H(g) ⎯→ H+(g) : ΔH4 = 1310 kJ / mol

Cl(g) ⎯→ Cl(g) : ΔH5 = – 348 kJ / mol

Cl2(g) ⎯→ 2Cl (g) : ΔH6 = 124 kJ / mol

N2(g) + 2H2(g) + Cl2(g) ⎯→ NH4(s) ΔH7 = – 314 kJ / mol

ΔH = (ΔH1) – ΔH2 + ΔH3ΔH4ΔH5(ΔH6) + ΔH7

= –× 218 + 46 + 683 – 1310 + 348 – × 124 – 314  

= – 718 kJ / mol

Problem 11: Calculate the enthalpy change when infinitely dilute solutions of CaCl2 and Na2CO3 mixed. for Ca2+ (aq), (aq) and CaCO3(s) are – 129.80,
– 161.65,  – 288.5 kcal mole–1 respectively.

Solution: On mixing CaCl2 (aq) and Na2CO3

CaCl2 + Na2CO3 ⎯→ CaCO3 + 2NaCl

Solutions are very dilute and thus 100% dissociation occurs 

Ca2+(aq)+2Cl(aq)+2Na+ (aq) +(aq) CaCO3+2Na+(aq)+2Cl (aq) or  Ca2+(aq)+ + CO32– (aq) CaCO3(s)

ΔH = Σproducts  – Σreactants 

or ΔH = 

ΔH° of a compound = ΔH° formation  = –288.5 – (–129.8 – 161.65)

= 2.95 kcal

Problem 12: Standard heat of formation of CH4, CO2 and H2O (l) are – 76.2, – 394.8 and
– 241.6 kJ mole–1. Calculate the amount of heat evolved by burning 1m3 of CH4 measured under normal conditions.

Solution: Given C+ 2H2 CH4 ΔH = – 76.2 kJ ———- (1)

C + O2 CO2 ΔH = – 394.8 kJ ——— (2)

H2 + 1/2 O2 H2O (l) ΔH = – 241.6 kJ ———- (3)

Multiplying equation (3) by 2 and adding in equation (2) 

C+ 2H2 + 2O2 CO2 + 2H2O (g) ΔH = – 878 kJ ———- (4) 

Subtracting equation (1) from equation (4) we have 

CH4 + 2O2 CO2 + 2H2O (g) ΔH = –801.8 kJ

Since 22.4 litres or 22.4×10–3 m3 CH4 on burning gives energy = 801.8 kJ

1m3 of CH4 on burning will  give = 35.79 ×  103 kJ

Heat evolved by burning 1m3 CH4 is 35.79 × 103 kJ

Problem 13: Calculate the heat of reaction for the hydrogenation of acetylene to ethylene at constant volume at 25°C from the following data 

  1. i) Enthalpy of  formation of water = –285.8 kJ mol–1
  1. Enthalpy of combustion of acetylene = –1299.6 kJ mol–1
  2. Enthalpy of combustion of ethylene = –1410.8 kJ mol–1

Solution: i) H2(g) + 1/2  O2(g) ⎯→ H2O (l) , ΔH = –285.8 kJ mol–1

  1. ii) C2H2(g) + O2 (g) ⎯→ 2CO2 (g) + H2O (l) , ΔH = –1299.6 kJ mol–1

iii) C2H4(g) + 3O2 (g) ⎯→ 2CO2(g) + 2H2O (l), ΔH = –1410.8 kJ mol–1

Eqn. (ii) + Eqn (i) – Eqn (iii), we get 

C2H2(g) + H2(g) ⎯→ C2H4(g)

ΔH = –1299.6 –285.8 + 1410.8 

= –174.6 kJ mol–1

ΔE = –174.6 –(–1) (8.314 × 10–3) (298) 

= –172.12 kJ mol–1 

Problem 14: In order to get maximum calorific output a burner should have an optimum fuel to oxygen ratio which corresponds to 3 times as much oxygen as is required theoretically for complete combustion of fuel. A burner which has been adjusted for methane as fuel (with x litre/hour of CH4 and 6 x  litre/hour of O2) is to be readjusted for butane C4H10. In order to get same calorific output, what should be the rate of supply of butane and oxygen. Assume that losses due to incomplete combustion etc., are the same for both fuels and that gases behave ideally. Heats of combustion:CH4 = 809 kJ/mol; C4H10 = 2878 kJ/mol.

Solution: CH4 + 2O2 ⎯→ CO2 + 2H2O; ΔH = –809 kJ mol–1

Initial volume/hr x 6x   x   6x

(in litre)

Let the temperature be T and assume volume of 1 mole of a gas is V litre at this condition

V litre or 1 mole CH4 gives energy on combustion = 809 kJ

x litre of CH4 gives energy on combustion = kJ

2878 kJ energy is obtained by 1 mole or V litre C4H10

kJ energy is obtained by = litre C4H10

= 0.281 (x) litre C4H10

Thus butane supplied for same calorific output = 0.281 x litre

C4H10 + ⎯→ 4CO2 + 5H2O; ΔH = –2878 kJ / mole.

Volume of O2 required = 3 × Volume of O2 for combustion of C4H10

= 3 × × volume of C4H10 = 3 × × 0.281 (x) 

= 5.48(x) litre O2

Problem 15: When 12.0 g of carbon reacted with oxygen to form CO and CO2 at 25°C and constant pressure, 75.0 kcal of heat was liberated and no carbon remained. Calculate the mass of oxygen needed for it and moles of CO and CO2 formed. Given ΔHf CO2 = –94.05 and ΔHf CO = –26.41 kcal mole–1

Solution: Here C + O2 CO2; ΔH = –94.05 kcal

C + ½ O2 CO; ΔH = –26.41 kcal

Let a mole of carbon reacts to form CO2 and b moles of carbon reacts to form CO.

Since 12g or 1 mole carbon is used up

a + b = = 1 (1)

a × 94.05 + b× 26.41 = 75 (2)

a = 0.718 i.e. mole of CO2 formed

b = 0.282 i.e. mole of CO formed

Also moles of O2 used for this change = 0.718 + = 0.859

Thus weight of O2 used = 0.859 × 32 = 27.49 gm

11.2 Objective 

Problem 1: (1) For the given heat of reaction,

(i) C(s) + O2(g) = CO2(g) + 97 kcal

(ii) CO2(g) + C(s) = 2CO(g) – 39 kcal

the heat of combustion of CO(g) is: 

(A) 68 kcal (B) – 68 kcal

(C) +48 kcal (D) None 

Solution: Substracting equation (ii)  from equation (i), we get 

C(s) + O2(g) = CO2(g) + 97 kcal

CO2(g) + C(s) = 2CO(g) – 39 kcal

————————————————————————

or, –CO2(g) + O2(g) = CO2(g) – 2CO(g) + 136 kcal

or, 2CO(g) + O2 = 2CO2(g) + 136 kcal

or, CO(g) + 1/2 O2(g) = CO2(g) + 68 kcal

Required value  = 68 kcal

(A)

Problem 2: For the combustion of n – octane

C8H18 + O2 ⎯→ CO2 + H2O at 25°C (ignoring resonance in CO2)

(A) ΔH = ΔE – 5.5 × 8.31 × 0.298 in  kJ/mol

(B) ΔH = ΔE + 4.5 × 8.31 × 0.298 in kJ/mol

(C) ΔH = ΔE – 4.5 × 8.31 × 0.298 in kJ/mol

(D) ΔH = ΔE – 4.5 + 8.31 × 0.298 in kJ/mol

Solution: C8H18(l) + 12.5 O2(g) ⎯→ 8CO2(g) + 9H2O(l)

ΔH = ΔE + (8 – 12.5) × 8.31 × 0.298 in kJ/mol

(C)

Problem 3: The enthalpies of neutralisation of HClO4 and Cl3C – COOH are

– 13.5 kcal/g – equivalent and

– 13.5 kcal/g – equivalent respectively

when 40 gm of solid NaOH is added to a mix of 1 gm mol. HclO4 and 1 g mol. Cl3C – COOH, sodium perchlorate and sodium trichloroacetate are formed in molar ratio of 3:1. Then.

(A) ΔH for the reaction of NaOH with the mix. Is 6.45 kcal.

(B) ΔH for the reaction of NaOH with the mix. Is 13.8 kcal.

(C) After reaction the total no. of moles of acid remained are 0.5 (HclO4 & Cl3C – COOH)

(D) After reaction the total wt. of acid remained is 147.75 gm. (HclO4 + Cl3Cl – COOH) and ΔH for the reaction of NaOH with mix. is –13.8 kcal.

Solution: HclO4 + NaOH ⎯→ NaClO4 + H2O

Cl3C – COOH + NaOH ⎯→ Cl3C – COONa + H2O

Let x mol of NaOH are used in 1st reaction and (1 – x) mol of NaOH are used in 2nd reaction.

Given

ΔH = –13.5 × + (–14.7) ×

= = –13.8 kcal

Weak acids = (1 + 35.5 + 64) + 3 (35.5 × 3 + 24 + 32 + 1)

= 147.75 gm

Problem 4: When 40 gm. NaOH reacts with a mix. Of 1 mol. each of two weak mono protic acids HX and HY it is found that 75% of HX is neutralized. If the value of Ka of HX is 9 × 10–6 then

  1. A) the pH of 0.01 M solution of HY in water is 4 and degree of ionisation of 0.01 M HY in water is 0.1%.
  2. B) The pH of 1 M solution of HY in water is 3 and degree of ionisation of 1 M HY in water is 0.1%.
  3. C) Ka (HY) = 3 × 10–6
  4. D) Ka (HY) = × 10–6

Solution: HX + OH ⎯→ X + H2O

HY + OH ⎯→ Y + H2O

Since 75% of HX is neutralised

i.e. 0.75 mol of HX is neutralised

0.25 mol of HY will be neutralised

Apply Ka =

9 × 10–6 = [H+] =

Ka (HY) = [H+] ×

× 3 = 9 Ka (HY) = 1 × 10–6

[H+] =

pH = 3

α = = 10–3 i.e. 0.1%

Problem  5: In the reaction AB2(l) + X2(g) AX2(g) + BX2(g)

ΔH = –270 kcal per mol. of AB2(l)

The enthalpies of formation of

AX2(g) & BX2(g) are in the ratio of 4:3 and have opposite sign.

The value of (AB2(l)) = +30 kcal/mol. Then

(A) (AX2) = –96 kcal/mol

(B) (BX2) = + 480 kcal/mol

(C) Kp = Kc & (AX2) = + 480 kcal/ mol

(D) Kp = Kc RT & (AX2) + (BX2) = 840 kcal/mol

Solution: Let (AX2) = + α

and (BX2) = β

Given β = 0.75α

Now AB2(l) + 3X2(g) AX2(g) + 2BX2(g)

–270 = α + 2 (–β) – (30)

–240 = α – 2β

–240 = α = 1.5α

α = 480 kcal/mol

(AX2) = 480 kcal/mol

(BX2) = –360 kcal/mol

(C)

Problem  6. In which case of mixing of a strong acid and a base each of 1 N concentration, temperature increase is highest.

(A) 20 ml acid – 30 ml alkali (B) 10 ml acid – 40 ml alkali

(C) 25 ml acid – 25 ml alkali (D) 35 ml acid – 15 ml alkali

Solution: Highest amount of heat will be released when equivalents of acid and base are equal i.e., complete neutralisation takes place.

(C)

Problem  7: The heat of combustion of carbon is 394 kJ. The heat evolved in combustion of 6.023 × 1022 atoms of carbon is

(A) 3940 kJ (B) 394.0 kJ

(C) 39.4 kJ (D) 0.394 kJ

Solution: Heat of combustion = × 6.023 × 1022

= 39.4 kJ  (C)

Problem 8: The enthalpy of combustion of H2(g) at 298 K to give H2O is –298 kJ mol–1 and bond enthalpies of H–H and O=O are 433 kJ mol–1 and 492 kJ mol–1 respectively. The bond enthalpies of O–H is 

(A) 464 kJ mol–1 (B) 488.5 kJ mol–1

(C) 232 kJ mol–1 (D) –232 kJ mol–1

Solution: H2 + 1/2O2 ⎯→ H2O + ΔH = – 298 kJ mol–1

ΔH = – [2ΔHO–H]

– 298 =  – [2ΔHO–H]

= ΔHO–H =

= 488.5 kJ /mol

(B)

Problem  9. The heat evolved in combustion if 112 litres of water gas (mixture of equal volume of H2(g) and CO(g) is

H2(g) + 1/2 O2(g) = H2O(g) ΔH = –241.8 kJ

CO(g) + 1/2  O2(g) = CO2(g) ΔH = –283 kJ

(A) 241.8 kJ (B) 296.5 kJ

(C) 1312 kJ (D) 1586 kJ

Solution: On adding equation (1) and (2), we get  

H2 + CO + O2 ⎯→ H2O + CO2 + ΔH = – 524.8 kJ

2 × 22.4 litre (H2 + CO)  evolve 524.8 kJ heat 

112 litre water gas evolved = × 112

= 1312 kJ

(C)

Problem 10: If H2 + 1/2  O2 H2O, ΔH = –68.39 kcal

K + H2O KOH (aq) + 1.2 H2, ΔH = –48 kcal

KOH + water KOH (aq), ΔH = –14 kcal

The heat of formation of KOH is

(A) – 68.39 + 48 – 14 (B) – 68.39 – 48 + 14

(C) + 68.39 – 48 + 14 (D) + 68.39 + 48 – 14

Solution: Eq. (i) + Eq. (ii) – Eq.(iii), we get  the desired equation and
ΔH = ΔH1 + ΔH2ΔH3

= – 68.39 – 48 + 14

(B)

Problem 11: For the combustion reaction at 298 K

H2(g) + 1/2 O2(g) H2O (l)

Which of the following alternative(s) is/are correct

(A) ΔH = ΔE

  1. ΔH > ΔE
  2. ΔH < ΔE

(D) ΔH &  ΔE has no relation with each there

Solution: Δn = –1/2 –1

= – 3/2

Since, Δn < 0

ΔH is < ΔE

(C)

Problem 12: The difference between the heat of reaction of constant pressure and constant volume for the reaction 

2C6H6(l) + 15O2(g) 12CO2(g) + 6H2O(l) at 25°C in kJ is

(A) –7.43 (B) 3.72

(C) –3.72 (D) 7.43

Solution: For the reaction 

2C6H6(l) + 15O2(g) ⎯→ 12CO2(g) + 6H2O(l)

Here Δn = 12–15 = –3 

As we know,

ΔH = ΔE + ΔnRT

or, ΔH – ΔE = ΔnRT

= – 3 × kJ = – 7.43 kJ 

(A)

Problem 13: The heat of neutralisation of HCl by NaOH is –12.1 kJ/mole, the energy of dissociation of HCN is

(A) –43.8 kJ (B) 45.4 kJ

(C) 68 kJ (D) –68 kJ

Solution: Since some heat is used up to dissociate weak acid, HCN

Let the enthalpy of ionization or dissociation of weak acid = x 

As we know that 

ΔH = neutralisation of strong acid and strong base  = – 57.5 kJ / eq.

– 57.5 = –12.1 –x 

= 45.4 kJ /  equivalent 

(B)

Problem 14: The dissociation energy of CH4 and C2H6 are respectively 360 & 620 k. cal/mole. The bond energy of C–C is 

(A) 260 kcal/mole (B) 180 kcal/mole

(C) 130 kcal/mole (D) 80 kcal/mole

Solution: 4ΔHC–H = 360 

ΔHC–H = = 90 KCal / mole

From question for the dissociation of C2H6

Energy needed is 620 Kcal / mole 

i.e, ΔHC–C + 6ΔHC–H = 620

ΔHC–C + 6 × 90 = 620 

∴ΔHC–C = 620 – 540 = 80 Kcal / mole 

(D)

Problem 15: The heat of combustion of propane C3H8(g) is ΔH = –2280 kJ /mol. For the combustion of 1.00 m3 of C3H8(g) measured at 25° C annd 1 atm pressure, ΔH =

  1. –2220 kJ
  2. + (1.00/22.4) × 2220 kJ
  3. –(1.00 / 22.4) × (298 /273) × 2220 kJ
  4. 1000/(0.082 × 298) × 2280 kJ

Solution: n =

Combustion of 1 mole of propane releases 2280 kJ heat 

Combustion of mole of propane releases 

× 2280 kJ 

(D)

Problem 16: The heat of combustion of CS2 is –3.24 kcal. For the reaction, 

CS2(l) + 3O2 CO2(g) + 2SO2(g)

(A) –3.24 kcal (B) –1.03 × 103 kJ

(C) –58.95 kJ (D) –13.6 kJ

Solution: Enthalpy of combustion = – 3.24 Kcal / gm 

Enthalpy of combustion 

Per mole of CS2 = – 3.24 KCal × 76 

= – 3.24 ×  4.18 × 76 

= – 1029.28 kJ 

= – 1.03 × 103 kJ

(B)

Problem 17: If the enthalpy of vaporistion for water is 186.5 kJ/mol, its entropy of
vaporistion will be 

(A) 0.5 kJ mol–1K–1 (B)1.0 kJ K–1 mol–1

(C) 1.5 kJ K–1 mol–1 (D) 2.0 kJ K–1 mol-1

Solution: Since 

ΔS =  

=

= 0.5 KJ / mol K

(A)

Problem 18: If ΔG0 > 0, for a reaction then:

(A) Kp > 1

(B) Kp < 1

(C) The products predominate in the equilibrium mixture

(D) None

Solution: ΔG0 = 2.303 RT log Kp

(B)

Problem 19: Highest entropy is in

(A) Water (B) Hydrogen

(C) Mercury (D) Graphite

Solution: Randomness or disoder is maximum in gaseous phase.

∴ (B)

Problem 20: Which is not a spontaneous process

(A) Expansion of a gas into vaccum

(B) Water flowing down hill

(C) Heat flowing from colder body to a hotter body

(D) Evaporation of water from clothes during drying

Solution: Heat cannot be it self pass from colder to hotter body

(C)

 

  1. Assignment (Subjective Problems)

 

LEVEL – I

  1. The heat of combustion of Acetylene is 312 kcals. If the heat of formation of CO2 is 94.38 kcals and that of water is 68.38 kcals. Calculate the C C bond energy; assuming that the bond energy of C – H is 93.6 kcals. Heat of atomisation of C and H are 150 and 51.5 kcals respectively.
  2. The heat of solution of anhydrous CuSO4 is – 15.9 kcal and that of CuSO4.5H2O is 2.8 Kcal. Calculate the heat of hydration of CuSO4.
  3. Heats of neutralisation (ΔH) of NH4OH and HF are –51.5 and 68.6 kJ respectively. Calculate their heats of dissociation.
  4. Calculate the heat of reaction at 25°C for the reaction:

C2H4(g) + H2(g) ⎯→ C3H6(g)

Given that the heats of combustion of ethylene, hydrogen and ethane at 25°C are 337.0, 68.4 and 373.0 kcal respectively.

  1. The commercial production of water gas utilizes the reaction under standard conditions: C + H2O(g) ⎯→ H2 + CO. The heat required for this endothermic reaction may be supplied by adding a limited amount of air and burning some carbon to CO2. How many gm of carbon must be burnt to CO2 to provide enough heat for the water gas conversion of 100g carbon. Neglect all heat losses to environment. of CO, H2O(g) and CO2 are –110.53, –241.81 and –393.51 kJ/mol respectively.
  2. Heat of combustion of glucose at constant pressure at 17°C was found to be
    –651,000 cal. Calculate the heat of combustion of glucose at constant volume considering water to be in the gaseous state.
  3. The combustion of methane takes place according to the following equation:

CH4(g) + 2O2(g) ⎯→ CO2(g) + 2H2O(l); ΔH = –8903 kJ

  1. a) How many grams of methane would be required to produce 89.03 kJ of heat of combustion?
  2. b) How many grams of carbon dioxide would be formed when 89.03 kJ of heat is evolved?
  3. A Carnot’s engine working between two temperatures has an efficiency of 40%. When the temperature of the sink is reduced by 60°C, the efficiency increases to 55%. Calculate the two temperatures in the first case.
  4. 0.50 g of benzoic acid was subjected to combustion in a bomb calorimeter at 15°C when the temperature of the calorimeter system (including water) was found to rise by 0.55°C. Calculate the heat of combustion of benzoic acid: (i) at constant volume, (ii) at constant pressure. The thermal capacity of the calorimeter system including water was found to be 23.85 kJ.
  5. The enthalpy change involved in the oxidation of glucose is –2880 kJ mol–1. Twenty five percent of this energy is available for muscular work. If 100 kJ of muscular work is needed to walk one kilometre, what is the maximum distance that a person will be able to walk after eating 120 g of glucose?
  6. The heat of reaction (ΔH) for the formation of NH3 according to the reaction
    N2 + 3H2 2NH3 at 27°C was found to be –91.94 kJ kJ. What will be the heat of reaction ΔH at 50°C. R [= 8.31 J/mol/0k].
  7. For the reaction A2(g) + 3B(g) ⎯→ AB(g) + AB2(g)

Calculate Given molar heat capacity of monoatomic gas = (α + βT) J/mol

diatomic gas =

polyatomic gas = (α, β are constatns)

  1. A piston exerting a pressure of 1.0 atm rests on the surface of water at 100°C. The pressure is reduced to smaller extent and as a result 10g of water evaporates and absorbs 22.2 kJ of heat. Calculate (a)  ΔH (b) heat of vaporisation  ΔHV (c) W (D) ΔE.
  2. From the following data, calculate  the enthalpy change for the combustion of cyclopropane at 298 K. The enthalpy of formation of CO2(g), H2O(l) and propene (g) are – 393.5, – 285.8 and 20.42 kJ mol–1 respectively. The enthalpy of isomerisation of cyclopropane to propene is – 33.0 kJ mol–1.    
  3. Calculate the heat of formation of KOH (s) using the following equations.

K(s) + H2O(l) + aq ⎯→ KOH (aq) + 1/2 H2(g) ΔH = – 48 kcal …(1)

H2(g) + 1/2 O2 (g) ⎯→ H2O(l) ΔH = – 68.4 kcal …(2)

KOH (s) + aq ⎯→ KOH (aq) ΔH = –14.0 kcal …(3)

 

LEVEL – II

  1. Isothermally at 27°C, one mole of a vander waal’s gas expands reversibly from 2 litres to 20 litres. Calculate the work done if a  =1.42 × 1012 dynes cm4 per mole and b = 30 c.c
  2. The bond enthalpy of H2(g)  is 436 kJ mol–1 and that of N2(g) is 941.3 kJ
    mol–1. Calculate the average bond enthalpy of an N – H bond in ammonia (NH3) = –46.0 kJ mol–1.
  3. Calculate the heat of neutralization from the following data 200 ml of 1M HCl is mixed with 400 ml of 0.5 M NaOH. The temperature rise in calorimeter was found to be 4.4°C. Water equivalent of calorimeter is 12 g and specific heat is 1 cal / ml / degree for solution.
  4. 2g of ethylalcohol is burnt in excess of O2 in bomb calorimeter and the products analysed are H2O(l) and CO2(g) and unreacted O2. If the heat capacity of the system is 1600 JK–1 and the rise in temperature is 37.1 deg, calculate the heat evolved in this reaction.
  5. The heats of formation of PCl3 and PH3 are 306 kJ mole–1 and +8kJ mole–1 respectively, and the heats of atomization of phosphorus, chlorine and hydrogen are given by:

Ps = Pg ΔH = 314 kJ mole–1

Cl2(.g) = 2Clg ΔH = 242 kJ mole–1

H2,(g) = 2Hg  ΔH = 433 kJ mole–1

Calculate EP – Cl  and EP – H

  1. 0.16 g of methane was subjected to combustion at 27°in a bomb calorimeter. The temperature of calorimeter system (including water) was found to rise by 0.5°C. Calculate the heat of combustion of methane at (i) constant volume (ii) constant pressure. The thermal capacity of the calorimeter system is 17.7kJ K–1.
  2. The heat of formation of one mole of HI from hydrogen and iodine vapour at 25°C is 8000 cal.

H2(g) + I2(g) ⎯→ HI(g), ΔH = – 8000 cal

Calculate the heat of formation at 10°C and also the total change in the heat capacity at constant pressure. It is given that the molar heat capacities of hydrogen, iodine and HI vapours are given by the equations.

CP = 6.5 + 0.0017 T for Hydrogen (g)

CP = 6.5 + 0.0013 T for Iodine (g)

Cp = 6.5 + 0.0016 T for HI (g)

Where T is the absolute temperature

  1. An athelete takes 20 breaths / min at room temperature. The air inhaled in each breath is 20 ml which contains 20% O2 by volume, while exhaled air contains 10% oxygen by volume. Assuming that all the oxygen consumed if is used for converting glucose into CO2 and H2O(l), how much glucose will be burnt in the body in 1 hour and what is the heat produced. (Room temperature = 27°C and enthalpy of combustion of glucose is – 2822.5 kJ mole–1 at 0°C).
  2. Water is brought to boil under a pressure of 1.0 atm. When an electric current of 0.50 A from a 12 V supply is passed for 300 s through a resistance in thermal contact with it, it is found that 0.789g of water is vaporised. Calculate the molar internal energy and enthalpy changes at boiling point (373.15 K).
  3. Determine the enthalpy of formation of anhyd. Al2Cl6(s)

Given

2Al(s) + 6HCl(aq.) ⎯→ Al2Cl6(aq.) + 3H2(g) + 239760 cal.

ΔHf (HCl(g)) = –22 kcal/mol

Maximum amount of heat of solution of HCl(g) in water = –17315 cal/mol

Maximum amount of heat of solution of Al2Cl6(s) in water = –153.69 kcal/mol.

*11. The heat of neutralization of (i) CHCl2 – COOH by NaOH is 12830 cazl. (ii) HCl by NaOH is 13680 cal. (iii) NH4OH by HCl is 12270 cal. What is the heat of neutralization of dichlioro acetic acid by NH4OH? Calculate also the heats of ionization of dichloro acetic acid and NH4OH.

  1. Calculate the resonance energy of N2O from the following data. 

of N2O = 82 kJ mole–1.

Bond energies N N 946 kJ mole–1

N = N 418 kJ mole–1

O = O 498 kJ mole–1

N = O 607 kJ mole–1

  1. A constant pressure calorimeter consists of an insulated beaker of mass 92 g made up of glass with heat capacity 0.75JK–1g–1. The beaker contains 100 ml of 1 M HCl at 22.6°C to which 100 ml 1 M NaOH at 23.4°C is added. The final temperature after the reaction is complete is 29.3°C. What is ΔH per mole for this neutralization reaction. Assume that the heat capacities of all solutions are equal to that of same volumes of water.
  2. The heat of combustion of glycogen is about 476 kJ/mole of carbon. Assume that average rate of heat loss by an adult male is 150 watts. If we were to assume that all the heat comes from oxidation of glycogen, how many units of glycogen (1 mole carbon per unit) must be oxidised per day to provide for this heat loss.

*15. At 25°C, the following heat of formation are given:

Compound SO2(g) H2O(l)

kJ/.mole –296.81 –285.83

For the reactions at 25°C

2H2S(g) + Fe(s) FeS2(s) + 2H2(g) ΔH° = –137 kJ /mole

H2S(g) + 3/2O2(g) H2O(l) + SO2(g) ΔH° = –562 kJ/mole

Calculate heat of formation of H2S(g) and FeS2(s) at 25°C

 

LEVEL – III

  1. Consider the classroom of 5m × 10m × 3m. Initially t = 20°C and P = 1 atm. There are 50 people in class, each loosing energy to the room at average rate = 150 watt. Assume that walls, ceiling, floor and furniture are perfectly insulated and do not absorb any heat. How long will physical chemistry examination last if the professor has foolishly agreed to dismiss the class when air temperature in the room reaches body temperature 37°C. For air Cp =
  2. The integral enthalpy of solution of one mole of H2SO4 in the amount n mole of water is given by the equation ΔH =

Calculate ΔH for following processes.

  1. a) 1 mole of H2SO4 dissolved in 2 mole of H2O
  2. b) 1 mole of H2SO4 dissolved in 7 mole of H2O
  3. c) solution (a) + 5 mole of H2O
  4. d) 1 mole of H2SO4 dissolved in excess of H2O
  5. Calculate the electron affinity of Br from following data.

Lattice energy of NaBr = – 736 kJ mole–1 B.E.(Br – Br) = 192 kJ mole–1

ΔHf0 (NaBr) = – 376 kJ mole–1 (I.E.)NA = 490 kJ mole–1

ΔH298K of sublimation of Na = 109 kJ mole–1

  1. At 1000K from the data, N2(g) + 3H2(g) ⎯→ 2NH3(g)

ΔH0 = – 123.77 kJ mole–1

Substance N2 H2 NH3

Cp/R 3.5 3.5 4

Calculate the heat of formation of NH3 at 300K.

  1. A slice of banana weighing 2.502 g was burnt in a bomb calorimeter and produced a temperature rise of 3.05°. In the same calorimeter combustion of 0.316g sample of C6H5CO2H produced a temperature rise of 3.24°. The heat of combustion of benzoic acid at constant volume is – 3227 kJ mole–1. If an average banana weighs 12g how many cals can be obtained from one average banana.
  2. The enthalpy of formation of H2O(l) is – 285.77 kJ mole–1 and enthalpy of neutralisation of a strong acid and strong base is – 56.07 kJ. What is the enthalpy of formation g OH ions?
  3. A Carnot’s engine works between 120°C and 30°C. Calculate the efficiency. If the power produced by the engine is 400 watts, calculate the heat absorbed from the source and rejected to the sink every second.
  4. The heats of formation of carbon dioxide and water are 395 kJ and 285 kJ respectively, and the heat of combustion of acetic acid is 869 kJ (all exothermic). Calculate the heat of formation of acetic acid.
  5. The enthalpy of evaporation of water at 373 K is 40.67 kJ mol–1. What will be the enthalpy of evaporation at 353 K and 393 K if average molar heats at constant pressure in this range for water in liquid and vapour states are 75.312 and 33.89 JK–1 mol–1 respectively?
  6. At 25°C, the following heat of formation are given:

Compound SO2(g) H2O(l)

kJ/.mole –296.81 –285.83

For the reactions at 25°C

2H2S(g) + Fe(s) FeS2(s) + 2H2(g) ΔH° = –137 kJ /mole

H2S(g) + 3/2O2(g) H2O(l) + SO2(g) ΔH° = –562 kJ/mole

Calculate heat of formation of H2S(g) and FeS2(s) at 25°C

  1. Compute the heat of reaction at 10000 C for

H2(g) +Cl2(g) ⎯→ HCl(g) ΔH0298K = 92.236 kJ mol–1

(H2, g = 29.0284 – 0.8355 × 10–3T + 2.0097 × 10–6 T2

(Cl2, g) = 31.6555 + 1.0134 × 10–2 T – 4.0337 × 10–6 T2

(HCl, g) = 28.1359 + 1.8078 × 10–3 T + 1.5453 × 10–6 T2

  1. From the following data:

ΔHC – H = 416 kJ/mol ΔHO – O = 494 kJ/mol

ΔHC = O – 711 kJ/mol ΔHO – H = 464 kJ/mol

ΔHH – H = 435 kJ/mol

ΔHCl – Cl = 243 kJ/mol RE (CO2) = 181 kJ/mol

Calculate ΔH0comb (CH4),

(CH3Cl(g), (H2O(l))

and then calculate ΔH° for CH4(g) + Cl2(g) ⎯→ CH3Cl(g) + HCl(g)

Also calculate ΔH° for the monochlorination of methane without finding heats of combustion of CH4(g), CH3Cl(g) and heat of formation of H2O(l) and HCl(g)

  1. An intimate mixture of Fe2O3 and Al is used in solid fuel rocket. Calculate the fuel value per gram and per ml of mixture. = 399.0 Kcal.  = 199.0 kcal density of Fe2O3 and Al are 5.2 g / ml and 2.7 g / ml respectively.
  2. Assume that for a domestic hot water supply 150 kg of water per day must be heated from 10°C to 65°C and gaseous fuel propane is used for this purpose. What moles and volume of propane in litres at STP would have to be used for heating domestic water. ΔH of combustion of propane is – 2050 kJ and specfic heat of water is 4.18 × 10–3 kJ / g 
  3. Compute the heat of formation of liquid methyl alcohol in kilojoules per mole, using the following data. Heat of vaporization of liquid methyl  alcohol = 38 kJ / mole . Heat of formation of gaseous atoms from their elements in their standard states:
    H = 218 kJ / mole, C = 715 kJ / mole , O = 219 kJ . mole . Average bond energies:
    C–H = 415 kJ / mole, C–O = 356 kJ / mole , O – H =463 kJ / mole

 

  1. Assignment (Objective Problems)

 

LEVEL – I

In each of the objective question, FOUR alternative(s) are given of which one or more than one is/are correct(s). Select the correct alternative(s).

  1. Work done in reversible isothermal process is given by

(A) 2.303 RT log (B)

(C) 2.303 RT log (D) None

  1. Work done in reversible adiabatic process is given by

(A) 2.303 RT log (B) None

(C) 2.303 RT log (D)

  1. Entropy change of vaporisation at constant pressure is given by

(A) ΔS(v) = (B) ΔS(v) =

(C) ΔS(v) = (D) None

  1. The work done during the process when 1 mole of gas is allowed to expand freely into vaccum is

(A) 0 (B) +ve

(C) –ve (D) Either of these

  1. 1 mole of an ideal gas at 25°C is subjected to expand reversibly ten times of its initial volume. The change in entropy of expansion is

(A) 19.15 JK–1 mol–1 (B) 16.15 KJ–1 mol

(C) 22.,15 JK–1 mol–1 (D) None

  1. The entropy change for the reaction given below is

2H2(g) + O2(g) ⎯→ 2H2O(l)

is … at 300 K. Standard entropies of H2(g). O2(g) and H2O9l) are 126.6, 201.20 and 68.0 JK–1 mol–1 respectively.

(A) –318.4 JK–1 mol–1 (B) 318.4 × JK–1 mol–1

(C) 31.84 × JK–1 mol–1 (D) None

  1. The enthalpy of neutralisation of the given reaction

H2SO4 + 2NaOH Na2SO4 + 2H2O + y kcal is

(A) y kcal (B) –y kcal

(C) + y/2 kcal (D) –y/2 kcal

  1. When solute remains in equilibrium with given solvent then

(A) ΔHhydration = Lattice energy (B) ΔHhydration < Lattice energy

(C) ΔHhydration > Lattice energy (D) None

  1. At  equilibrium condition, the value of Gibbs free energy change, ΔG is

(A) Equal to zero (B) Greater than one

(C) Less than one (D) Equal to one

  1. For spontaneous reaction the value of change of Gibbs free energy, ΔG is

(A) Negative (B) Positive

(C) Greater than one (D) One

  1. For isochoric process, the change in

(A) Volume of zero (B) Pressure of zero

(C) Internal energy is zero (D) None

  1. Which of the following is the heat of combustioin?
  1. C(graphite) + 1/2  O2(g) CO(g) + x cal
  2. C(diamond) + O2(g) CO2(g) + y cal
  3. C(diamond) + 1/2  O2(g) CO(g) + z cal
  4. None
  1. Heat of reaction of constant p or constant v varies with temperature as given by Kirchoff’s equation is/are

(A) ΔH2 = ΔH1 + ΔCP (T2– T1) (B) ΔE2 = ΔE1 + ΔCP (T2 – T1)

(C) H2 = H1 + ΔCp (T2 – T1) (D) ΔE2 = ΔE1 + Cp (T2 – T1)

  1. For the combustion reaction at 298 K

H2(g) + 1/2 O2(g) H2O (l)

Which of the following alternative(s) is/are correct

(A) ΔH = ΔE

(B) ΔH > ΔE

(C) ΔH < ΔE

(D) ΔH &  ΔE has no relation with each there

  1. The temperature of a 5 ml of strong acid increases by 5°C when 5 mol of a strong base is added to it. If 10 ml of each are mixed, temperature should increase by

(A) 5°C (B) 10°C

(C) 15°C (D) Cannot be known

  1. If H2(g) = 2H(g); ΔH = 104 cal, then heat of atomisation of hydrogen is

(A) 52 kcal (B) 104 kcal

(C) 208 kcal (D) None of these

  1. Given N2(g) + 3H(g) = 2NH3(g); ΔH0 = –22 kcal. The standard enthalpy of formation of NH3 gas is

(A) –11 kcal/mol (B) 11 kcal/mol

(C) –22 kcal/mol (D) 22 kcal/mol

  1. Heat of combustion of CH4, C2H4, C2H6 are –890, –1411 and –1560 kJ/mol respectively. Which has the lowest calorific fuel value in kJ/g.

(A) CH4 (B) C2H4

(C) C2H6 (D) All same

  1. ΔH for CaCO3(s) ⎯→ CaO(s) + CO2(g) is 176 kJ mol–1 at 1240 K. The ΔE for the change is equal to

(A) 160 kJ (B) 165.6 kJ

(C) 186.3 kJ (D) 180.0 kJ

  1. Cdiamond + O2(g) ⎯→ CO2(g); ΔH = –395 kJ ..(i)

Cgraphite + O2(g) ⎯→ CO2(g); ΔH = –393.5 kJ ..(ii)

The ΔH, when diamond is formed from graphite is

(A) –1.5 kJ (B) +1.5 kJ

(C) +3.0 kJ (D) –3.0 kJ

 

LEVEL – II

  1. The temperature at which the reaction,

Ag2O(s) ⎯→ 2Ag(s) + 1/2O2(g)

Is at equilibrium is …; Given ΔH = 30.5 kJ mol–1 and ΔS = 0.066 kJ K–1 mol–1

(A) 462.12 K (B) 362.12 K

(C) 262.12 K (D) 562.12 K

  1. If 50 calorie are added to a system and system does work of 30 calorie on surroundings, the change in internal energy of system is

(A) 20 cal (B) 50 cal

(C) 40 cal (D) 30 cal

  1. If S0 for H2, Cl2 and HCl are 0.13, 0.22 and 0.19 kJ K–1 mol–1 respectively. The total change in standard entropy for the reaction H2 + Cl2 ⎯→ 2HCl is

(A) 30 JK–1mol–1 (B) 40 JK–1mol–1

(C) 60 JK–1mol–1 (D) 20 JK–1mol–1

  1. If H+ + OH ⎯→ H2O + 13.7 kcal then the heat of neutralisation for complete neutralisation of 1 mole of H2SO4 by base will be

(A) 13.7 kcal (B) 27.4 kcal

(C) 6.85 kcal (D) 3.425 kcal

  1. Heat evolved in the reaction H2 + Cl2 ⎯→ 2HCl is 182 kJ. Bond energies of H – H and Cl – Cl are 430 and 242 kJ/mol respectively. The H – Cl bond energy is

(A) 245 kJ moil–1 (B) 427 kJ mol–1

(C) 336 kJ mol–1 (D) 154 kJ mol–1

  1. Heat or formation of H2O(g) at 1 atm and 25°C is –243 kJ. ΔE for the reaction H2(g) + 30 JK–1mol–1 O(g) ⎯→ H2O(g) at 25°C is

(A) 241.8 kJ (B) –241.8 kJ

(C) –243 kJ (D) 243 kJ

  1. The H – H bond energy is 430 kJ mol–1 and Cl – Cl bond energy is 240 kJ mol–1. ΔH for HCl is –90 kJ. The H – Cl bond energy is about.

(A) 425 kJ mol–1 (B) 213 kJ mol–1

(C) 360 kJ mol–1 (D) 180 kJ mol–1

  1. The enthalpies of combustion of carbon and carbon monoxide are –390 kJ and –278 kJ respectively. The enthalpy of formation of CO in kJ is

(A) 668 (B) 112

(C) –112 (D) –668

  1. Under the same conditions how many ml of 1 M KOH and 0.5 M H2SO4 solutions, respectively when mixed for a total volume of 100 ml produce the highest rise in temperature

(A) 67 : 33 (B) 33 : 67

(C) 40 : 60 (D) 50 : 50

  1. ΔH of sublimation of a solid is equal to

(A) ΔHfusion + ΔHcondensation (B) ΔHcondensation + ΔHsublimation

(C) ΔHcondensation ΔHfusion (D) ΔHvaporisation + ΔHfusion

  1. The expression, is true at all

(A)Temperatures

(B) Pressures

      (C) Temperatures & pressures

(D) Temperature &1atm pressure conditions

  1. According to the equation

C6H6(l) +15/2 O2(g) 3H2O (l) + 6 CO2(g)

ΔH = –3264.4 kJ mol–1, the energy evolved when 7.8 gm of benzene is burnt in air will be

(A) 163.22 kJ/mol–1 (B) 326.4 kJ/mol

(C) 32.64 kJ/mol–1 (D) 3.264 kJ/mol

  1. The product of combustion of an aliphatic thiol (RSH) at 298 K are

(A) CO2(g), H2O(g), & SO2(g) (B) CO2(g), H2O(g), H2O(l) & SO2(g)

(C) CO2(l), H2O(l) & SO2(g) (D) CO2(g), H2O(l), SO2(l)

  1. Given ΔHioniz (HCN) = 45.2 kJ mol–1 & ΔHioniz (CH3COOH) = 2.1 kJ/mol
  1. pKa (HCN) = pKa (CH3COOH)
  2. pKa (HCN) > pKa (CH3COOH)
  3. pKa (HCN) < pKa (CH3COOH)
  4. pKa (HCN) = pKa (CH3COOH)
  1. AB, A2 and B2 are diatomic molecules. If the bond enthalpies of A2, AB & B2 are `in the ratio 1:1:0.5 and enthalpy of formation of AB from A2 and B2 –100 kJ/mol–1. What is the bond enthalpy of A2?

(A) 400 kJ/mol (B) 200 kJ/mol

(C) 100 kJ/mol (D) 300 kJ/mol

  1. Which of the following corresponds to the definition of enthalpy of formation at
    298 K?
  1. C(graphite) + 2H2(g) + 1/2 O2(l) CH3OH(g)
  2. C(diamond) + 2H2(g) + 1/2 O2(g) CH3OH (l)
  3. 2C(graphite) + 4H2(g) + 4H2(g) + O2(g) 2CH3OH (l)
  4. C(graphite) + 2H2(g) + 1/2 O2(g) CH3OH(l)
  1. In which case, a reaction is possible at any temperature?

(A) ΔH < 0, ΔS > 0 (B) ΔH < 0, ΔS < 0

(C) ΔH > 0, ΔS > 0 (D) None

  1. Which of the following has impossibilities of reaction of any temperature?

(A) ΔH > 0, ΔS < 0 (B) ΔH > 0, ΔS > 0

(C) ΔH < 0, ΔS < 0 (D) None

  1. The temperature of 5 ml of a strong acid increases by 5° C when 5 ml of a strong base is added to it. If 10 ml of each is mixed, temperature should increase by

(A) 5° C (B) 10° C

(C) 15° C (D) Cannot be known

  1. Identify the intensive quantities from the following:

(A) Enthalpy (B) Temperature

(C) Volume (D) Refractive index

 

  1. Answers to Subjective Assignments

LEVEL – I

  1. 160.94 kcals 2. – 18.7 kcal
  2. – 11.3 kJ 4. – 32.3 kcal
  3. 33.36 gm 6. – 654, 480 cal
  4. a) 1.6 g 8. T1 = 400 K, T2= 240 K
  5. b) 4 g CO2
  6. – 3200 kJ, – 3201.9 kJ 10. 4.8 km
  7. – 93.087 kJ 12. (– 100 + 135 β – 0.42α) kJ
  8. a) 22.2 kJ 14. – 2091.31 kJ
  9. b) 56 15. – 102.4 kcal
  10. c) 723 kJ
  11. d) 477 kJ

LEVEL – II

  1. – 58124.67 Joules 2. 390.2 kJ mole–1
  2. – 13.464 kcal 4. – 1367.75 kJ mole–1
  3. P – Cl = 328 kJ 6. – 889.95 kJ/mole

P – H = 319.33 kJ 7. – 7995 cal

  1. 458.374 kJ 9. 40.6 kJ/mole, 37.5 kJ/mole
  2. – 321960 cal/mole 11. a) – 11420 cal
  3. – 88 kJ b) 850 cal/mole, 1410 cal/mole
  4. – 57 kJ 14. 27.22 units
  5. – 20.64 kJ/mole, – 178.28 kJ/mole

LEVEL – III

  1. 6.88 min 2. a) – 39.79 kJ
  2. – 335 kJ mole–1 b) – 60.14 kJ
  3. – 44.42 kJ mole–1 c) – 20.35 kJ
  4. – 93.59 kcals d) – 75.6 kJ
  5. – 229.7 kJ 7. 1747 watts, 1347 watts
  6. – 491.kJ 9. – 41.422 kJ and 39841.56 J
  7. – 20.64 kJ mole–1, – 178.28 kJ mole–1 11. – 94.827 kJ mole–1
  8. – 286 kJ mole–1, – 92 kJ mole–1, 13. 0.9346 kcal/g, 3.94 kcal/ml

– 102 kJ mole–1 14. 377.17 litres

  1. – 266 kJ mole–1

 

  1. Answers to Objective Assignments

 

LEVEL – I

  1. A 2. B
  2. A 4. A
  3. A 6. A
  4. D 8. A
  5. A 10. A
  6. A 12. D
  7. A, B 14. B
  8. A 16. A
  9. A 18. B
  10. B 20. B

LEVEL – II

  1. A 2.     
  2. A 4. B
  3. A 6. B
  4. A 8. C
  5. D 10. D
  6. D 12. B
  7. A 14. B
  8. A 16. D
  9. A 18. B
  10. A 20. A

Thermochemistry

 

Hints for Subjective Problems

 

Level – I

 

  1.     Apply Hess’s law
  2.     DH neutralization of strong acid and strong base is – 57.3 kJ/equivalent.
  3.     First calculate DHcombustion per mole of carbon then for given amount.
  4.     First calculate DH, then DE
  5.   Apply Kirchoff’s equation
  6.   Apply Hess’s law

 

Level – II

 

  1.     heat produced during neutralization will be due to calorimeter and solution.
  2.     O2 is completely consumed so acts as limiting reagent. O2 used only will produce heat by burning glucose
  3.   N2O have structure N  N = O.
  4.   For getting rise in temperature, get initial temperature by taking average.

 

Level – III

 

  1.     First calculate heat released by 50 people per second.
  2.     Apply Bron Haber cycle.
  3.     First get DCp then Kirchoff’s equation
  4.     Relate DT with heat produced in calorimeter
  5.     Apply Kirchoff’s equation
  6.   Use integrated form of Kirchoff’s equation
  7.   From heat change we can go for volume of propane.

  

Subjective Problems

 

Level – I

 

  1.     Heat of formation of Acetylene:
  2.   2C + H2 ¾® C2H2                         DH = ?
  3.   C2H5 + O2 ¾® 2CO2 + H2O     DH = – 312 kcals
  4.   C + O2 ¾® CO2                           DH = – 94.38 kcals
  5.   H2 + O2 ¾® H2O                      DH = – 68.38 kcals

         In order to arrive at (1), multiply (3) by 2, and add (4) to it and substract (2) from their sum so we get,

         2C + H2 ¾® C2H2                               DH = +54.86 kcals

         Formation of C2H2 involves atomisation of  2C atoms and absorbing 150 ´ 2 = 300 kcals and would release energy of the order of 93.6 ´ 2 = 187.2 kcals for the formation of two C – H bonds and also release energy for the formation of one C º C bond

(x kcals) the value of which can be worked out as under

         300 + 103 – 187.2 – x = 54.86

         x = 160.94 kcals

  1.     CuSO4 + aq ¾®  CuSO4.5H2O (aq)              DH = – 15.9 kcal ———– (1)

         CuSO4.5H2O + aq ¾® CuSO4.5H2O (aq)  DH = 2.8 kcal ————(2)

         Subtracting (2) from (1)                                  

         CuSO4 + 5H2O ¾® CuSO4.5H2O           DH = – 15.9 – 2.8

         = –18.7 kcal

         \Heat of hydration of CuSO4 = – 18.7 kcal

  1.     We have
  2. i) HCl (aq) + NaOH (aq) ¾® NaCl (aq) + H2O; DH = –57.3 kJ
  3. ii)   HCl (aq) + ¾® NH4Cl (aq) + H2O;  DH = –51.5 kJ

         \ The heat of dissociation of NH4OH

         D H = – 51.5 – (57.3) = 5.8 kJ

         Similarly we have

         HF (aq) + NaOH (aq) ¾® NaF + H2O;  DH = –68.6 kJ

         \ The heat of dissociation of HF,

         DH = –68.6 – (– 57.3) = –11.3 kJ

  1.     The thermochemical equations for the combustion of ethylene , hydrogen and ethane may be written as
  2. i) C2H4 + 3O2 ¾® 2CO2 + 2H2O   DH = –337 kcal
  3. ii)   H2 +O­2¾® H2O            DH = –68.4 kcal

         iii)   C2H6 + 3O2 ¾® 3CO2 + 3H2O DH = –373 kcal

         (i) + (ii) – (iii) Þ C2H4 + H2 – C2H6

         Heat of reaction = –337 – 68.4 + 373 = – 32.3 kcal.

  1.     C + H2O(g)  ¾¾® H2 + CO         = ?

         (Q DH°  for C and H2 are zero)

         \ DH°  = –110.53 – (–241.81) = 131.28 kJ/mol

         Thus, DH° needed for 100g carbon =

         Now 393.51 kJ energy is provided by 12g C

         kJ energy is provided by =  = 33.36 g

  1.    

         Dn = 12 – 6 = 6

         DH = DE + DnRT

         DE = DH – DnRT = –651000 – 6 ´ 2 ´ (273 + 17)

         = –651000 – 3480 = –654480 cal

         Hence heat of combustion of glucose at constant volume and at

         17°C is –654, 480 cal.

 

  1.     a)   Molecular weight of CH4 = 1´ 12 + 4 ´ 1 = 12 + 4 = 16

             890.3 kJ of energy is lilberated by the combustion of 1 mole of 16 g of methane.

               \ 89.03 kJ of energy is liberated by the combustion of 16 ´  = 1.6g of methane.

               1.6 g of methane

  1. b)   Molecular weight of CO2 = 1 ´ 12 + 2 ´ 16

   = 12 + 32 = 44

               [ C = 12, O = 16]

               890.3 kJ of energy is liberated to produce 1 mole or 44 g of CO2

               \ 89.03 kJ of energy is liberated to produce 44 ´  or 4.4 g of CO2

  1.     h = 0.4 =

         \

         When the temperature of the sink is reduced by 60°C, we get efficiency, h¢.

         h = 0.55 =

         =

         \

         On dividing equation (2) by (1), we obtain

        

         or 4T2 – 240 = 3T2

         T2= 240 K

         T1 =

9.     i)    Heat of combustion at Constant Volume (DE):

               Weight of benzoic acid, m = 0.50 g

               Rise in temperature,   q = 0.55°C

               Molecular weight of benzoic acid,  M = 122

               Thermal capacity of the calorimeter system, z = 23.85 kJ

           \ Heat of combustion at constant volume, DE = z ´ q ´

       = –23.85 ´ 0.55 ´  = – 3200.7 kJ = – 3200.7 kJ

  1. ii)     Heat of combustion at constant pressure (DH):

               The combustion of benzoic acid at 25°C involves the following reaction:

               C6H5COOH(s) + 7.5O2(g) ¾® 7CO2(g) + 3H2O(l)

               DH = DE + DnRT

               In the present case:

               DE = –3200.7 kJ Dn = 7 – 7.5 = –2.5         R = 8.314 ´ 10–3 kJ

               T = 273 + 25 = 298°K

               \ DH = –3200.7 + (–0.5) ´ 8.314 ´ 10–3 ´ 298 = –3201.9 kJ

               \ Heat of combustion at constant pressure, DH = –3201.9 kJ

  1.   Energy available for mascular work =  = 720 kJ / mole

         Energy evolved on tbe burning of 120 g glucose = = 480 kJ

         Distance travelled by using 480 kJ energy

         =  = 4.8 Km

  1.  

         (We know CP (diatomic gas) =

         \

        

  1.   Apply

         Þ d(DH) =

         Integrate

         Þ

        

         = (–100 ´ 103 + 300 {450 b – 14a})J

         = (– 100 + 0.3 ´ 450 b – 0.42 a) kJ = (–100 + 135 b – 0.42 a) kJ

  1.   H2O(e) H2O(g)
  2. a)   DH = 22.2 kJ = heat absorbed
  3. b)   mol of water vapourised = n = = 0.56

               \ latent heat of vapourisation DHV =  = 39.96 kJ mole–1

  1. c)   W = nRT

               = 0.56 ´ 8.314 ´ 10–3 ´ 373.15

               = 1.723 kJ

  1. d)   DE = DH – W = 22.2 – 1.723

               = 20.477 kJ

  1.   C(s) + O2 ¾® CO2(g) DH1 = – 393.5 kJ ———- (1)

         H2(g) +  O2(g) ¾¾® H2O(l)   DH2 = – 285.8 kJ ————– (2)

         3C(s) + 3H2(g) ¾¾® C3H6    DH3 = 20.42 kJ —————– (3)

    

         DH = 3DH1 + 3DH2 – DH3 + DH4

         = 3 ´ – 393.5 + 3 ´ (– 285.3) – 20.42 – 33

         = – 2089. 82 kJ

  1.   We have to evaluate DH for

         K(s) + O2(g) + H2(g) ¾® KOH (s)

         Adding equation (1) and (2), we have

     K(s) + H2(g) + O2(g) + aq ¾® KOH (aq)  DH = –116.4 kcal ——–(4)                                    Subtracting (3)  from (4) we have.

     K(s) + 1/2 H2(g) + 1/2 O2(g) ¾®  KOH (s)  

         DH = – 102.4 kcal

 

Level – II

 

  1.     W  =

         =  = RTln +

         = 2.303 ´ 8.31´107 ´ 300 log  + 1.42 ´ 1012 

         = – 58124.67 ´ 107 ergs

         = – 58124.67 Joules

  1.       i) N2 (g) ¾® 2N (g);        DH0 = 941.3 kJ mol–1
  2. ii)   H­2(g) ¾® 2H(g);  DH0 = 436.0 kJ mol–1

               iii)   N2(g) + 3/2H2(g) ¾® 2NH3(g); DH0‑ = –46.0 kJ mol–1

                     On multiplying equation (i) by and equation (ii) by 3/2 and

adding, we get

  1. iv)  N2(g) + 3/2H2(g) ¾® N(g) + 3H(g),    DH0 = 1124.6 kJ mol–1

                     On substracting equation (iii) from (iv), we get (v) NH3 ¾® N(g) + 3H(g)     DH° = 1170.6 kJ mole–1

                     As there are three N – H bonds, in NH3, the average bond enthalpy is obtained by dividing the value DH° from (v) by 3. Hence

                     DHN – H = 1170.6/3 = 390.2 kJ mol–1

  1.     Given total solution  =  200 + 400 = 600 ml

         200 ml of 1 M HCl neutralise 400 ml 0.5M NaOH

         Meq = N ´V

         \ Meq of acid and base = 200

         i.e., 200 Meq of HCl react with 200 Meq of NaOH to produce heat = DH.

         1000 Meq of HCl + 1000 Meq of NaOH will give heat = 5 ´ DH i.e, heat of neutralization.

         Now heat produced during neutralization of 200 Meq of acid and base = Heat taken up       by Calorimeter  + solution

         = m1,s1DT + m2s2DT

         = 12 ´ 4.4  + 600 ´ 1 ´ 4.4 =  2692.8 kcal

         \ Heat of neutralization = – 5´ 2692.8 cal = – 13.464 kcal

  1.     The chemical equation is

         C2H5OH(l) + 3O2(g) ¾® 2CO2(g) + 3H2O(l)

         So, Dn = – 1

         Heat evolved here is at constant volume, So

         DE = QV = C×DT = (1600 JK–1) (37.1 K)

         = 59360 J

         This heat is given by 2g of C2H5OH, so 1 mole will give,

         = 1365.28 kJ mole–1

         i.e. = DE = – 1365.28 kJ mole–1

         Using equation DH = DE + RTDn

         DH = (–1365.28) + (–1) (8.314 ´ 10–3) (298)

         DH = – 1367.75 kJ mole–1

  1.     For phosphorus trichloride, thermochemical equations are:

         Ps = Pg        DH1 = 314 kJ

         3/2 Cl2(g) = 3Cl(g)   DH­2 = 363 kJ

         Pg + 3Clg = PCl3   DH­3 = ?

         Also,

         Ps + 3/2Cl2.(g) = PCl3   DHf = –306 kJ

         Equating the two routes leading to the formation of PCl3

         DHf = DH1 + DH2 + DH3

         –306 = 314 + 363 + DH3

         DH3 = –983 kJ

         This is energy released on the formation of three P – Cl bonds, hence,

         EP – Cl =  = 328 kJ mole –1

         Ps = Pg        DH1 = 314 kJ

         3/2H2.(g) = 3Hg         DH2 = 649 kJ

         P2 + 3Hg = PH3 DH3= ?

         Also,

         Ps + 3/2 H2(.g) = PH3   DHf = 8 kJ

         From which

         DH3 = ––958 kJ

         EP – H  =  = 319.33 kJ

  1.     Heat evolved at constant volume  = 0.16 ´ 17.7 ´ 0.5 kJ

         Heat of combustion of methane (heat evolved per mole)

         =  = 885 kJ / mol

         DE = – 885 kJ / mol

         CH4(g) + 2O2(g) ¾® CO2(g) + 2H2O (l)

         Dn = 1–3 = – 2

         DH = DE + DnRT

         = – 885 + (–2) ´ 8.314 ´ 10–3 ´ 298

         DH  = – 889.95 kJ / mole

  1.     DCp = CP(HI)

         = 6.5 + 0.0016T –  [6.5 + 0.0017T + 6.5 + 0.0038T]

         = – 0.00115T cal. deg–1

         Here DH298 = DH283 +

         Where T1 = 273 + 10 = 283K, T2 = 273 + 25 = 298 K

         Also, DH293 K = –8000 cal.

         –8000 = DH283  +

         or DH283 = –8000 +

         = –7995 cal (nearly)

  1.     Volume of O2 used in one breath

         = Inhaled O2 – exhaled O2 =  = 20 ml

         Volume of O2 used in one hour  = 60 ´ 20 ´ 20  = 24000 ml

         = 24 litre

         Number of moles of O2 =

         =  = 0.9744

         C6H12O6(s) + 6O2 ¾® 6CO2 + 6H2O

         DH = – 2822.5 kJ

         Weight of glucose required by 0.9744 moles of O2

             = ´ 0.9744 = 29.23 gm

         heat produced = ´ 0.9744 ´ 2822.50 = 458.374 kJ

  1.     Since the vapourisation occurs at constant pressure the enthalpy change is equal to the work done on the heater (which enters the water as heat):

         DH = 0.50A ´ 12V ´ 300s = 1800J

         = + 1.8 kJ

         \ Molar enthalpy of vapouristaion DHm=

         =  = 40.6 kJ mole–1

         Also, DHm = DEm + PDV

         = DEm + DngRT

         = DEm + RT (Q H2O(l)  H2O(g) Dng = 1)

         \ DEm= molar internal energy change = DHm – RT

         = 40.6 – 8.314 ´ 10–3 ´ 373.15

         = 37.5 kJ mole–1

  1.   For 2Al(s) + 3Cl2(g) ¾® Al2Cl6(s) enthalpy change is DH  which is

         DHf (Al2Cl6)

         Now

         Ii)   2Al(s) + 6Hcl(aq.) ¾® Al2C6(aq.) + 3H2(g)    DH1 = –239760 cal

  1. ii)   H2(g) + Cl2(g) ¾® 2HCl(g) DH2 = –44000 cal

         iii)   HCl(g) + aq. ¾® HCl(aq.)   DH3 = – 17315 cal

  1. iv)  Al2Cl6(s) + aq. ¾® Al2Cl6(aq.) DH4 = –153690 cal

         Apply (i) + 3 ´ (ii) – (iv) + 6 ´ (iii)

         DHf = –239760 + 3 (–44000) – ( – 153690) + 6(– 17315)

         = –321960 cal/mol

  1.   i) CHCl2COOH + OH ¾® CHCl2COO + H2O   DH1 = 12830 cal.
  2. ii)   H+ + OH ¾® H2O      DH2 = 13680 cal.

         iii)   NH4OH + H+¾®  + H2O DH3 = –12270 cal

         Consider

iii)    CHCl2COOH +NH4OH ¾® CHCl2COO +  + H2O

  1. iv)  Will be obtained by applying (i) + (iii – (ii)

         DH (for iv) = DH1 + DH3 – DH2

         = –12830 – 12270 + 13680

         = –11420 cal

         Consider now

  1. v)   CHCl2COOH ¾® CHCl2COO + H+
  2. vi)  will be obtained by applying (i) – (ii)

         \DH [for (v)] = DH1 = DH2

         = 13680 – 12830

         = 850 cal/mol

         Similarly DHionisatioin for NH4OH

         = 13680 – 12270

         = 1410 cal/mol

  1.   N º N  + O2 ¾®  = O

         The resonating structures of N2O is as follows

         N º N ® O ¬¾®  = O

         (i)                             (ii)

         Structure (ii) is more acceptable

         The value of DH for the above reaction  is given by

         DH = SBEreactants – SBEproducts

         DH =   – (418 + 607)   = 1195 – 1025

         = 170 kJ mole

         Resonance Energy  = (DHf)experimental – (DHf)Theoretical  = 82 – 170

         = – 88 kJ

  1.   Initial average temperature of the acid and base

         =  = 23.0°C

Rise in temperature = (29.3 – 23.0) = 6.3°C

Total heat produced = (92 ´ 0.75 + 200 ´ 4.184) ´ 6.3    =  5706.54 J

Enthalpy of neutralization

= = –57065.4 J = –57 kJ

  1.   Total energy required in the day =

= 12960 kJ        [Q  Watt = J sec–1]

Units of glycogen required =  = 27.22 units

  1.   Heat of formation of H2S = x kJ/mole

Heat of formation of FeS2 = y kJ/mole

2H2S(g) + Fe(s) ® FeS2(s) + 2H2(g)   DH° = –137 kJ/mole

– 137 = y – 2x   ———–(1)

H2S(g) + O2(g) ® H2O(l)  + SO2(g)   DH° = –562/mole

–562 = –285.83 – 296.81 – x    ———–(2)

from equation (2), x = –20.64 kJ mole–1

from equation (1), we get, y = –178.28 kJ mole–1

 

LEVEL – III

 

  1.     Volume of air inside the room = V = 5 ´ 10 ´ 3 = 150 m3

         Mole of air at 298.15 K = n =  = 6.236 ´ 103

         DH = Cp(DT) ´ n (for n mole)

         = 6.236 ´ 103 ´ ´ 8.314 ´ 17 = 3.085 ´ 106 J

         Heat released by 50 people = 150 ´ 50 watt = 7500 watt = 7500 J/sec.

         Hence no. of seconds for which the examination last =  

         = 6.86 min

  1.     a)   DHa (n = 2) = –kJ = – 39.79kJ
  2. b)   DHb (n = 7) = kJ = – 60.14 kJ
  3. c)   For this, we are to determine integral heat of solution when solution containing one mole H2SO4 in 2 mole H2O is diluted and a solution containing 7 mole H2 Thus

               DHc = DHb – DHa = – 20.35 kJ

  1. d)   DHa =  =  =  = – 75.6 kJ
  2.     Reaction Na(s) + Br2(g) ¾® NaBr(s)   DH10 = – 376 kJ mole–1
  3. a)   Na(s) ¾® Na(g)                      DH0(sublimation) = S = 104 kJ mole–1
  4. b)   Na(g) ¾® Na+(g) + e                    DH0(ionisation) = I = 490 kJ mole–1
  5. c)   Br2(g) ¾® Br(g)                   DH0(dissociation) =  = 96 kJ mole–1
  6. d)   Br(g) + e ¾® Br(g)               DH0(electron affinity) = – E
  7. e)   Na+(g) + Br(g) ¾® NaBr             DH0(lattice energy) = – U = – 736 kJ mole–1
  8. f) Na(s) + Br2(g) ¾® NaBr(s) DH20 = – 376 kJ mole–1

         DH20 = S + I +  – E – U

         or Putting all the values, E = 335 kJ mole–1

         Electron affinity = – 335 kJ mole–1

  1.     Using Kirchoff’s equation

         DH2 (1000 K) = DH1 (300 K) + DCp (1000 – 300)

         DCp = 2Cp (NH3) – Cp(N2)  – 3C­p(H2)

         = – 6R = – 6 ´ 8.314 ´ 103 kJ

         \ – 123.77 = DH1 (300 K) – 6 ´ 8.314 ´ 103 ´ 700

         or, DH1 (300 K) = – 123.77 + 34.92 = 88.85 kJ for 2 mole of NH3

         DHf(NH3) = – 44.42 kJ mole‑1

  1.     –3227 kJ mole–1

         \  (for 0.316g) = kJ mole–1

         which increases temperature of 3.24°

         for 3.05° temperature rise

         DE =  kJ mole–1 for 2.502 g of banana

         \ single banana gives

         =  ´  kJ

         =  ´ kcals

         = – 93.59 kcals

  1.     DHf(H+, aq) = 0

         H+ + OH ¾® DH = 56.07 kJ

         – 56.07 = DHf(H2O) = DHf(H+) – DHf(OH)

         DHf(OH) = + 56.07 – 285.77 = – 229.7 kJ

  1.     Efficiency, h =

         Here T1 = 273 + 120 = 393 K

         T2 = 273 + 30 = 303 K

         \ h =  = 0.229 = 22.9%

         Again h =

         Thus, heat absorbed from the source,

         Q1 =

         = 1747 watts.

         Also

         \ The heat rejected to the sink

         Q2 = Q1 ´

         = 1747 ´  = 1347 watts

  1.     The data given are expressed in the form of three thermochemical equations.

         C + O2 = CO2   DH1 = –395 kJ (1)

         H2 + O2 = H2O  DH2 = –285 kJ (2)

         CH3COOH + 2O2 = 2CO2 + 2H2O DH3 = –869 kJ (3)

         The equation corresponding to the formation of acetic acid from its elements is

         2C + 2H2 + O2 ¾® CH3COOH  DHf = ?

         This equation may be obtained from the three thermochemical equations on multiplying equations (1) and (2) by two and adding, finally substracting equation (3):

         2 ´ (1)                2C + 2O2 = 2CO2

         2 ´ (2)                2H2 + O2 = 2H2O

         ––––––––––––––––––––––––––––––––––––––––––––––––––

         Adding               2C + 3O2 + 2H = 2CO2 + 2H2O

         Substract (3)           CH3COOH + 2O2 = 2CO2 + 2H2O

         ––––––––––––––––––––––––––––––––––––––––––––––––––

         Giving, after re-arranging 2C + O2 + 2H2 = CH3COOH

         2 ´ (1)                2 ´ DH1 = –790 kJ

         2 ´ (2)                2 ´ DH2 = –570 kJ

         ––––––––––––––––––––––––––––––––––––––––––––––––––

         Adding               DH = –1360 kJ

         Substract (3)           DH3 = –869 kJ

         ––––––––––––––––––––––––––––––––––––––––––––––––––

         Giving                DHf = –2360 – (–869) = –491 kJ

 

  1.     H2O(l)  H2O(g)

     Enthalpy change at 373 K is DH1 = 40.67 kJ mol–1

         = 40670 J mol1

         DH2 – DH1 = DCp (T2 – T1)

         DH2 = DH1 + DC1 (T2 – T1)

         DCp =

         33.89 – 75.312 = – 41.422

         Enthalpy of evaporation at 353 K, DH

         = 40670 – 41.522 (353 – 373) = 40670 + 41.421 ´ 20 = 41498.44 J

         Similarly enthalpy of evaporation at 393 K

         = 40670 – 41.422 (393 – 373) =39841.56 J

  1.   Heat of formation of H2S = x kJ/mole

Heat of formation of FeS2 = y kJ/mole

2H2S(g) + Fe(s) ® FeS2(s) + 2H2(g)   DH° = –137 kJ/mole

– 137 = y – 2x   ———–(1)

H2S(g) + O2(g) ® H2O(l)  + SO2(g)   DH° = –562/mole

–562 = –285.83 – 296.81 – x    ———–(2)

from equation (2), x = –20.64 kJ mole–1

from equation (1), we get, y = –178.28 kJ mole–1

  1.   DH01273K = DH0298 K +

Where DCp =  (HCl, g) –  –  (Cl2, g)

= (28.1359 – ´ 29.0284 – ´ 31`.6555) + (1.8078 +  ´ 0.8355 –  ´ 10.134) ´ 10–3 T

+ (1.5453 –  ´ 2.0097 + ´ 4.0337) ´ 10–6 T2

Hence,

DH01273K =

+ J mol–1

= –92236 – 2151.0 + 2176.2 + 1736.0 = –94827 J mol–1 = – 94.827 kJ mol–1

  1.   CH4 + 2O2 ¾® CO2(g) + 2H2OI(l)

         (CH4)m = 4 ´ 416 + 2 ´ 494 – {2 ´ 711 + 181 + 2 (40 + 2 ´ 464)}

         = {1664 + 988} – {1603 + 1936}

         = 2652 – 3539 = – 887 kJ/mol

         CH3Cl(g) + O2(g) ¾® CO2 + HCl + H2O(l)

         (CH3Cl) + (3 ´ 416 + 330) + (494)

         {2 ´ 711 + 181 + 431 + 2 ´ 464 + 40}

         = 1248 + 330 + 741 – {1422 + 181 + 431 + 928 + 40}

         = 2319 – 3002

         = –683 kJ/mol

         H2(g) + O2(g) ¾® H2O(l)

 ¾® H2O(l)

 = 435 +(494) – {2 ´ 464 + 40}

= 682 – 968 = – 286 kJ/mol

 +  ¾® HCl(g)

 (HCl(g)) =

= –92 kJ/mol

 

Now

  1. i) CH4 + 2O2 ¾® CO2 + 2H2O
  2. ii)   CH3Cl + O2 ¾® CO2 + HCl + H2O

iii)   H2 + O2 ¾® H2O

  1. iv)  H2 + Cl2 ¾® HCl
  2. iv)  CH4 + Cl2 ¾® CH3Cl + HCl
  3. v)   will be obtained by applying (i) + 2 ´ (iv) – (ii) – (iii)

\ DH [for (v)] = DH=1 + 2DH4 – DH2 – DH3

= – 887 + 2 ´ (-92) – (–683) – (–286)

= –102 kJ/mol

  1.   2Al + Fe2O3 ¾® 2Fe + Al2O3 + 2Fe   DH = ?

         Given,

     2Al + O2 ¾® Al2O3                         DH = – 399 kcal      …(1)

         2Fe + O2 ¾® Fe2O3                 DH = – 199.0 kcal         …(2)

         (2) – (1)

         2Al + Fe2O3 ¾® 2Fe +  Al2O3

         Fuel mixture                                    DH = – 200 kcal

         Molecular weight of Fuel mixture = 214g

         Q 214g produces 200 kcal heat

         1 g produces  = 0.9346 kcal/g

         Also volume of fusion mixture = col. Of Al + vol. of Fe2O3

         =  = 50.76 ml

         Q 50.76 ml produces 200 kcal heat

         1 ml produces  = 3.94 kcal /ml

  1.   Heat required to change the temperature of water from 10°C to 65°C

         = 150 ´ 103 ´ 4.184 ´ 10–3 ´ 55

         = 34518 kJ

         Number of moles of propane required to produce 34518 kJ heat

         =  = 16.838

         Volume of propane required (at NTP) = 22.4 ´ 16.838 = 377.17 Litres

  1.   C(g) + 4H(g) + O(g) ¾® CH3 – OH DHf = ?

         D[DHC(s)  ¾®  C(g) + 2DHH–H + 1/2 DHO = O]

         – [3DHC–H + DHC–O + DHO–H + DHVap(CH3OH(l))]

         = [ 715 + 2 ´ 436 + 249] – [3 ´ 415 + 356 + 463 + 38]

         = 1836 – 2102

         = – 266 kJ mol–1

                            

   Objective Problems

 

Level – I

  1.     (A) w = Pext (V2 – V1)

         Q Pext = 0

         Q w = 0

  1.     (A) DS =

         = 2.303 ´ 1 ´ 8.314 log 10

         = 19.15 JK–1 mol–1

  1.     (A) DsREACTION = åsproduct – åS­reactant

         = 2 ´  –

         = 2 ´ 68 – [2 ´ 126.6 + 201.20]

  1.   (A) No doubt heat evolved during neutralization of 10 ml of each acid and base is twice the heat evolved during neutralization of 5 ml of each acid and base but the quantity of solution taking heat is also doubled thus same temperature rise is noticed.
  2.   (A) Heat of atomisation of hydrogen is derived for 1 mole of atoms of hydrogen formed.
  3.   (A) Standard enthalpy of formation of NH3 is derived for 1 mole of NH3 formed.
  4.   (B) Calorific value = Heat of combustion per g of fuel, i.e. for C2H4, it is  , the lowest value.

 

  1.   (B) DH = DE + DnRT

         \DE = 176 – 1 ´ 8.314 ´ 8.314 ´ 1240 ´ 10–3

         = 165.6 kJ

 

  1.   (B) (ii) – (i)

         CC ¾® CD; DH = + 1.5 kJ

 

Level – II

  1.     (A) DG = DH = TDS

         DG = 0 at equilibrium

         \ DH = TDS

         or 30.5 = T ´ 0.66

         T = 462.12 K

  1.     (A) q = DE + w

         50 = DE + 30

         \ DE = 20 cal

  1.     (A) DS = SP – SR

         = (2 ´ 0.19) – 0.13 – 0.22

         = 0.03 kJ–1 mol–1 = 30 JK–1 mol–1

  1.     (B) 1 mole or 2 eq. of H2SO4 are neutralised.
  2.     (A) DH = –2 ´ eH – Cl + eH – H + eCl – C

         \ n182 =  –2 ´ a + 430 + 242

         \ a  = 245 kJ mol–1

  1.     (B) DH = DE + DnRT

         Dn = –1/2

         \ –43 = DE + (– 1/2) ´ 8.314 ´ 298 ´ 10–3

         \ DE = –241.76

  1.     (A) H2 + Cl2 ¾® HCl

         DH = –90 kJ

         \ DH = EH–H + E­Cl – Cl

         or – 90 =  ´ 430 +  ´ 240 – eHCl

         \ EH – Cl = 425 kJ mol–1

 

  1.     (C) C + O2 ¾® CO2; DH = –290 kJ               …(1)

         CO  + O2 ¾® CO2; DH = –278 kJ        …(2)

         By (1) and (2)

         C + O2 ¾® CO; DH = –112 kJ

  1.     (D) 50 Meq. of KOH and 50 Meq. of H2SO4 will produce maximum heat.

 

 

 

 

 

 

 

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